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<h2 class="hd hd-2 unit-title">Non-Measurable Sets</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> In this lecture we will see that there are subsets of [mathjaxinline]\mathbb{R}[/mathjaxinline] that are <strong>non-measurable</strong> in an especially strong sense. Not only do they fall outside the class of Lebesgue Measureable sets, they also cannot be assigned a measure by any <em>extension</em> of the LebesgueMeasure function [mathjaxinline]\lambda[/mathjaxinline], unless one gives up on one of Non-Negativity, Countable Additivity and Uniformity. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">(What does it mean for a function [mathjaxinline]\lambda^+[/mathjaxinline] to extend [mathjaxinline]\lambda[/mathjaxinline]? It means that whenever [mathjaxinline]\lambda(A)[/mathjaxinline] is defined, [mathjaxinline]\lambda^+(A)[/mathjaxinline] is also defined and equal to [mathjaxinline]\lambda(A)[/mathjaxinline]. For technical reasons, we will impose the additional constraint that [mathjaxinline]\lambda^+[/mathjaxinline] only counts as an extension of [mathjaxinline]\lambda[/mathjaxinline] if it is closed under intersections; in other words: [mathjaxinline]\lambda^+(A \cap B)[/mathjaxinline] must be defined whenever [mathjaxinline]\lambda^+(A)[/mathjaxinline] and [mathjaxinline]\lambda^+(B)[/mathjaxinline] are both defined.) </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">To get a sense of why the existence of sets that are non-measurable in this strong sense is such a striking result, suppose that [mathjaxinline]A[/mathjaxinline] is a subset of [mathjaxinline][0,1][/mathjaxinline] that is non-measurable. Now suppose that the Standard Coin-Toss Procedure of Lecture 7.1.4.1 is used to randomly select a point from [mathjaxinline][0,1][/mathjaxinline]. What is the probability that the selected point will fall within [mathjaxinline]A[/mathjaxinline]? The fact that [mathjaxinline]A[/mathjaxinline] is non-measurable means that <em>there is no such thing as the probability that the selected point will turn out to be in</em> [mathjaxinline]A[/mathjaxinline]</span></p>
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<h3 class="hd hd-2">Video Review: Brace Yourselves!</h3>
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<h2 class="hd hd-2 unit-title">The Axiom of Choice</h2>
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<p><span style="font-family: 'book antiqua', palatino;">In proving that there are non-measurable sets, we will use of the Axiom of Choice. (In fact, it is impossible to prove that there are non-measurable sets without some version of the Axiom of Choice. This was proved in 1970 by the American set theorist Robert Solovay.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The Axiom of Choice is based on a very simple idea. Suppose that <span class="math inline">\(A\)</span> is a set each element of which is also a set. Then a <strong>choice set</strong> for <span class="math inline">\(A\)</span> is a set that contains exactly one element from each element of <span class="math inline">\(A\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Suppose, for example, that <span class="math inline">\(A\)</span> is the following set: <span class="math display">\[\{\{0,1\},\{a,b,c\},\{3\},\{e,\pi,\iota\}\}\]</span> Then <span class="math inline">\(\{1, a, 3, \pi\}\)</span> is a choice set for <span class="math inline">\(A\)</span> (and, of course, there are many others). </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The <strong>Axiom of Choice</strong> states that every set has a choice set, as long as each of its members is a non-empty set and as long as no two of its members have any elements in common. More concisely:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Axiom of Choice</strong></span><br /><span style="font-family: 'book antiqua', palatino;">Every set of non-empty, non-overlapping sets has a choice set.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Upon first inspection, the Axiom of Choice is liable to sound trivial. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">But it is most certainly not. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The great British philosopher Bertrand Russell once suggested a nice way of bringing out the reason that the Axiom of Choice is so powerful. Let <span class="math inline">\(S\)</span> be a set consisting of infinitely many pairs of shoes, and suppose that you’d like to specify a choice set for <span class="math inline">\(S\)</span>. Since each pair of shoes consists of a left shoe and a right shoe, one way of doing so is to use “left shoe” as a selection criterion. Accordingly, one specifies a choice set by taking it to consist of all and only left shoes in members of <span class="math inline">\(S\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now consider a variation of the case. Let <span class="math inline">\(S'\)</span> be a set consisting of infinitely many pairs of <em>socks</em>, and consider the question of how one might go about specifying a choice set for <span class="math inline">\(S'\)</span>. Assuming there is no difference between “left socks” and “right socks”, it is not clear how one might go about doing so. For it is not clear that we are in a position to specify a criterion that would apply to exactly one member of each of our infinitely many pairs of socks: a criterion that might play the same role that “left shoe” played in specifying a choice set for <span class="math inline">\(S\)</span>. But the Axiom of Choice assures us that a choice set for <span class="math inline">\(S'\)</span> will exist regardless of our ability to specify such a criterion. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">That is the reason that Axiom of Choice is so interesting: it allows us to work with sets whose membership we are not in a position to specify. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">We’ll return to the Axiom of Choice in section 7.2.3.3.</span></p>
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<h3 class="hd hd-2">Video Review: The Axiom of Choice</h3>
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<h2 class="hd hd-2 unit-title">The Vitali Sets</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> The most famous examples of non-measurable sets are the Vitali sets, named in honor of the Italian mathematician Giuseppe Vitali, in 1905. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Vitali proved that there is no way of extending [mathjaxinline]\lambda[/mathjaxinline] to apply to a Vitali set while preserving all three of Non-Negativity, Countable Additivity and Uniformity. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">I’ll start by giving you an intuitive sense of why the Vitali sets are not measurable, and then go through the proof.</span></p>
<h4><span style="font-family: 'book antiqua', palatino;">The Intuitive Picture</span></h4>
<p><span style="font-family: 'book antiqua', palatino;">In Lecture 6 we considered a thought experiment: imagine that God has selected a positive integer, but that you have no idea which. What should your credence be that God selected the number 17? More generally: what should your credence be that She selected a given number [mathjaxinline]k[/mathjaxinline]? </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Our discussion revealed that as long as you think that your credence distribution ought to be <em>uniform</em>—as long as you think that the same answer should be given for each value of [mathjaxinline]k[/mathjaxinline]</span><span style="font-family: 'book antiqua', palatino;"> — </span><span style="font-family: 'book antiqua', palatino; font-size: 1em;">it is a consequence of Countable Additivity that your credences must remain undefined.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">For if [mathjaxinline]p[/mathjaxinline](God selects [mathjaxinline]k[/mathjaxinline]) [mathjaxinline]= 0[/mathjaxinline] for each [mathjaxinline]k[/mathjaxinline], Countable Additivity entails that [mathjaxinline]p[/mathjaxinline](God selects a positive integer) [mathjaxinline] = 0[/mathjaxinline], which is incorrect. And if [mathjaxinline]p[/mathjaxinline](God selects [mathjaxinline]k[/mathjaxinline]) [mathjaxinline]= r[/mathjaxinline] for each [mathjaxinline]k[/mathjaxinline], where [mathjaxinline]r[/mathjaxinline] is a positive real number, Countable Additivity entails that [mathjaxinline]p[/mathjaxinline](God selects a positive integer) [mathjaxinline]= \infty[/mathjaxinline], which is also incorrect. (We also saw that there are reasons of principle why appealing to infinitesimals won’t save the day, but we won’t have to worry about infinitesimals here, since we’re presupposing Non-Negativity.) </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The moral of our thought experiment is that in the presence of Countable Additivity, there is no such thing as a uniform probability distribution over a countably infinite set of (mutually exclusive and jointly exhaustive) possibilities.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The proof of Vitali’s Theorem is a version of this same idea. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">We’ll partition [mathjaxinline][0,1)[/mathjaxinline] into a countable infinity of “Vitali Sets”, and use Uniformity to show that these sets must all have the same measure, if they have a measure at all. We’ll then use Countable Additivity and Non-Negativity to show that the Vitali Sets cannot have a measure.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">When I described the thought experiment of Lecture 6, I didn’t say anything about the selection procedure God uses to pick a positive integer. In particular, I didn’t clarify whether there could be a selection procedure that doesn't have the feature that some numbers are more likely to be selected than others. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">You can think of Vitali’s Theorem as delivering one such procedure. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here's how it would work. God starts by partitioning [mathjaxinline][0,1)[/mathjaxinline] into a countable infinity of Vitali Sets, and assigns each member of the partition a distinct positive integer. (Proving that a partition of the right kind exists requires the Axiom of Choice, so God would have to rely on Her super-human capabilities to identify a suitable partition.) God then selects a real number in [mathjaxinline][0,1)[/mathjaxinline], using the Standard Coin-Toss Procedure of Lecture 7.1.4.1. Finally, God uses the real number she gets as output from the Coin-Toss Procedure to select the integer that corresponds to the member of the partition to which the real number belongs.</span></p>
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<h2 class="hd hd-2 unit-title">Proving the Theorem</h2>
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<h3 class="hd hd-2">Proof of the Vitali Theorem</h3>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">Here is a sketch the proof of the non-measurability of Vitali Sets. (Some of the technical details are assigned as exercises below.) </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">We'll start by <strong>partioning</strong> [mathjaxinline][0,1) [/mathjaxinline]. </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">In other words, we’ll divide [mathjaxinline][0,1)[/mathjaxinline] into a family of non-overlapping “cells”, whose union is [mathjaxinline][0,1)[/mathjaxinline]. </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">The cells are characterized as follows: for [mathjaxinline]a, b \in [0,1)[/mathjaxinline], [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] are in the same cell if and only if [mathjaxinline]a - b[/mathjaxinline] is a rational number. For instance, [mathjaxinline]\frac{1}{2}[/mathjaxinline] and [mathjaxinline]\frac{1}{6}[/mathjaxinline] are in the same cell because [mathjaxinline]\frac{1}{2} - \frac{1}{6} = \frac{2}{3}[/mathjaxinline], which is a rational number. Similarly, [mathjaxinline]\pi - 3[/mathjaxinline] and [mathjaxinline]\pi - \frac{25}{8}[/mathjaxinline] are in the same cell because [mathjaxinline](\pi - 3) - (\pi - \frac{25}{8}) = \frac{1}{8}[/mathjaxinline], which is a rational number. But [mathjaxinline]\pi - 3[/mathjaxinline] and [mathjaxinline]\frac{1}{2}[/mathjaxinline] are not in the same cell because [mathjaxinline]\pi - \frac{7}{2}[/mathjaxinline] is not a rational number. </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">We’ll call this partition of [mathjaxinline][0,1)[/mathjaxinline] [mathjaxinline]\mathcal{U}[/mathjaxinline], because it has uncountably many cells. The next step of our proof will be to use [mathjaxinline]\mathcal{U}[/mathjaxinline] to characterize a partition of [mathjaxinline][0,1)[/mathjaxinline] with countably many cells, which we’ll call [mathjaxinline]\mathcal{C}[/mathjaxinline]. Each cell of [mathjaxinline]\mathcal{C}[/mathjaxinline] will be a set [mathjaxinline]V_q[/mathjaxinline]for [mathjaxinline]q \in \mathbb{Q}^{[0,1)}[/mathjaxinline]. ([mathjaxinline]\mathbb{Q}^{[0,1)}[/mathjaxinline] is the set a rational numbers in [mathjaxinline][0,1)[/mathjaxinline].)</span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;"> I will now explain which elements of [mathjaxinline][0,1)[/mathjaxinline] to include in a given cell [mathjaxinline]V_q[/mathjaxinline] of [mathjaxinline]\mathcal{C}[/mathjaxinline]. </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">The first part of the process is to pick a representative from each cell in [mathjaxinline]\mathcal{U}[/mathjaxinline]. (In other words: we need a <em>choice set</em> for [mathjaxinline]\mathcal{U}[/mathjaxinline]. It is a consequence of Solovey’s result, mentioned above, that it is impossible to <em>define</em> a choice set for [mathjaxinline]\mathcal{U}[/mathjaxinline]. In other words: it is impossible to specify a criterion that that could be used to single out exactly one element from each cell in [mathjaxinline]\mathcal{U}[/mathjaxinline]. But it follows from the Axiom of Choice that a choice set for [mathjaxinline]\mathcal{U}[/mathjaxinline] must nonetheless exist. And its existence is all we need here, because all we’re aiming to show is that non-measurable sets exist.) </span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">We will now use our representatives from each cell in [mathjaxinline]\mathcal{U}[/mathjaxinline] to populate the cells of [mathjaxinline]\mathcal{C}[/mathjaxinline] with elements of [mathjaxinline][0,1)[/mathjaxinline]. The first step is to think of [mathjaxinline][0,1)[/mathjaxinline] as a line segment of length 1 (which is missing one of its endpoints), and bend it into a circle:</span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-14final2.png" type="saveimage" target="[object Object]" width="754" height="231" /><br /></span></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Recall that the difference between any two elements in a cell [mathjaxinline]\mathcal{U}[/mathjaxinline] is always a rational number. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">From this it follows that each member of [mathjaxinline][0,1)[/mathjaxinline] can be reached by starting at the representative of its cell in [mathjaxinline]\mathcal{U}[/mathjaxinline], and traveling some rational distance around the circle, going counter-clockwise. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Suppose, for example, that [mathjaxinline]\frac{3}{4}[/mathjaxinline] is in our choice set for [mathjaxinline]\mathcal{U}[/mathjaxinline], and has therefore been selected as the representative of its cell in [mathjaxinline]\mathcal{U}[/mathjaxinline]. (Call this cell [mathjaxinline]C_{\frac{3}{4}}[/mathjaxinline].) Now consider a second point in [mathjaxinline]C_{\frac{3}{4}}[/mathjaxinline]: as it might be, [mathjaxinline]\frac{1}{4}[/mathjaxinline]. Since [mathjaxinline]\frac{3}{4}[/mathjaxinline] and [mathjaxinline]\frac{1}{4}[/mathjaxinline] are in the same cell of [mathjaxinline]\mathcal{U}[/mathjaxinline], one can reach [mathjaxinline]\frac{1}{4}[/mathjaxinline] by starting at [mathjaxinline]\frac{3}{4}[/mathjaxinline], and traveling a rational distance around the circle, going counter-clockwise—in this case a distance of [mathjaxinline]\frac{1}{2}[/mathjaxinline]:</span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-15.png" alt="" type="saveimage" target="[object Object]" preventdefault="function (){r.isDefaultPrevented=n}" stoppropagation="function (){r.isPropagationStopped=n}" stopimmediatepropagation="function (){r.isImmediatePropagationStopped=n}" isdefaultprevented="function t(){return!1}" ispropagationstopped="function t(){return!1}" isimmediatepropagationstopped="function t(){return!1}" width="395" height="229" /></span></p>
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<p><span style="font-family: 'book antiqua', palatino;">If [mathjaxinline]a[/mathjaxinline] is point in [mathjaxinline][0,1)[/mathjaxinline], let us say that [mathjaxinline]\delta(a)[/mathjaxinline] is the distance one would have to travel on the circle, going counter-clockwise, to get to [mathjaxinline]a[/mathjaxinline] from the representative for [mathjaxinline]a[/mathjaxinline]'s cell in [mathjaxinline]\mathcal{U}[/mathjaxinline]. In our example, [mathjaxinline]\delta\left(\frac{1}{4}\right) = \frac{1}{2}[/mathjaxinline].</span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is now straightforward to explain how to populate the cells of our countable partition [mathjaxinline]\mathcal{C}[/mathjaxinline] with elements of [mathjaxinline][0,1)[/mathjaxinline]: each cell [mathjaxinline]V_q[/mathjaxinline] ([mathjaxinline]q \in \mathbb{Q}^{[0,1)}[/mathjaxinline]) of [mathjaxinline]\mathcal{C}[/mathjaxinline] is populated with those [mathjaxinline]a \in [0,1)[/mathjaxinline] such that [mathjaxinline]\delta(a) = q[/mathjaxinline]. As you’ll be asked to verify below, this definition guarantees that the [mathjaxinline]V_q[/mathjaxinline] ([mathjaxinline]q \in \mathbb{Q}^{[0,1)}[/mathjaxinline]) form a countable partition of [mathjaxinline][0,1)[/mathjaxinline].</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let a <strong>Vitali Set</strong> be a cell [mathjaxinline]V_q[/mathjaxinline] ([mathjaxinline]q \in \mathbb{Q}^{[0,1)}[/mathjaxinline]) of [mathjaxinline]\mathcal{C}[/mathjaxinline]. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">All that remains to complete our proof is to verify that the Vitali Sets must all have the same measure, if they have a measure at all. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The basic idea is straightforward. Recall that [mathjaxinline]V_q[/mathjaxinline] is the set of points at a distance of [mathjaxinline]q[/mathjaxinline] from their cell’s representative, going counter-clockwise. From this it follows that [mathjaxinline]V_q[/mathjaxinline] can be obtained by <em>rotating</em> [mathjaxinline]V_0[/mathjaxinline] on the circle counter-clockwise, by a distance of [mathjaxinline]q[/mathjaxinline]. So one can use Uniformity to show that [mathjaxinline]V_0[/mathjaxinline] and [mathjaxinline]V_q[/mathjaxinline] have the same measure, if they have a measure at all (and therefore that all Vitali Sets have the same measure, if they have a measure at all).</span></p>
<p><span style="font-family: 'book antiqua', palatino;">We are now in a position to wrap up our proof. We have seen that [mathjaxinline][0,1)[/mathjaxinline] can be partitioned into countably many Vitali Sets, and that these sets must all have the same measure, if they have a measure at all. But, for reasons rehearsed in Lecture 7.2.2.1, we know that in the presence of Non-Negativity and Countable Additivity there can be no such thing as uniform measure over a countable family of (mutually exclusive and jointly exhaustive) subsets of a set of measure 1. So there can be no way of expanding the notion of Lebesgue measure to Vitali sets, without giving up on Non-Negativity, Countable Additivity or Uniformity.</span></p>
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<span style="font-family: 'book antiqua', palatino;">The relation <span class="math inline">\(R\)</span>, which holds between <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> if and only if <span class="math inline">\(a - b\)</span> is a rational number, satisfies which of the following three properties?</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Reflexivity:</i> For every <span class="math inline">\(x\)</span> in <span class="math inline">\([0,1)\)</span>, <span class="math inline">\(xRx\)</span></span>
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<span style="font-family: 'book antiqua', palatino;"><i>Symmetry:</i> For every <span class="math inline">\(x\)</span> and <span class="math inline">\(y\)</span> in <span class="math inline">\([0,1)\)</span>, if <span class="math inline">\(xRy\)</span> then <span class="math inline">\(yRx\)</span>.</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Transitivity:</i> For every <span class="math inline">\(x\)</span>, <span class="math inline">\(y\)</span> and <span class="math inline">\(z\)</span> in <span class="math inline">\([0,1)\)</span>, if <span class="math inline">\(xRy\)</span> and <span class="math inline">\(yRz\)</span> then <span class="math inline">\(xRz\)</span>.</span>
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</div><span style="font-family: 'book antiqua', palatino;">(If </span>
<span class="math inline">\(R\)</span>
<span style="font-family: 'book antiqua', palatino;"> satisfies all three, then <span class="math inline">\(\mathcal{U}\)</span> is a partition of <span class="math inline">\([0,1)\)</span>.)</span>
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Problem 2
</h3>
<div class="problem-progress" id="block-v1:MITx+24.118x+2T2020+type@problem+block@7dda0433a5184c159c9ec91f47219911-problem-progress"></div>
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<div>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>
<span style="font-family: 'book antiqua', palatino;">Each cell of <span class="math inline">\(\mathcal{U}\)</span> has how many members?</span>
</p>
<div class="choicegroup capa_inputtype" id="inputtype_7dda0433a5184c159c9ec91f47219911_2_1">
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<input type="radio" name="input_7dda0433a5184c159c9ec91f47219911_2_1" id="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="7dda0433a5184c159c9ec91f47219911_2_1-choice_0-label" for="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_7dda0433a5184c159c9ec91f47219911_2_1">
<span style="font-family: 'book antiqua', palatino;">Countably many</span>
</label>
</div>
<div class="field">
<input type="radio" name="input_7dda0433a5184c159c9ec91f47219911_2_1" id="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="7dda0433a5184c159c9ec91f47219911_2_1-choice_1-label" for="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_7dda0433a5184c159c9ec91f47219911_2_1">
<span style="font-family: 'book antiqua', palatino;">Uncountably many</span>
</label>
</div>
<div class="field">
<input type="radio" name="input_7dda0433a5184c159c9ec91f47219911_2_1" id="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="7dda0433a5184c159c9ec91f47219911_2_1-choice_2-label" for="input_7dda0433a5184c159c9ec91f47219911_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_7dda0433a5184c159c9ec91f47219911_2_1">
<span style="font-family: 'book antiqua', palatino;">None of the above</span>
</label>
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<span id="answer_7dda0433a5184c159c9ec91f47219911_2_1"/>
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Problem 3
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>
<span style="font-family: 'book antiqua', palatino;"><span class="math inline">Show that \(\mathcal{U}\)</span> has uncountably many cells.</span>
</p>
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<span id="answer_9af69140f60a4cc59e8388e158eba0d4_2_1"/>
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Problem 4
</h3>
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<div class="problem">
<div>
<blockquote>
<p>
<span style="font-family: 'book antiqua', palatino;">Every real number in <span class="math inline">\([0,1)\)</span> belongs to some <span class="math inline">\(V_q\)</span> (<span class="math inline">\(q \in \mathbb{Q}^{[0,1)}\)</span>).</span>
</p>
</blockquote>
<p>
<span style="font-family: 'book antiqua', palatino;">True or false?</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_e50dfddf157442a193c59b4699ebe562_2_1">
<fieldset aria-describedby="status_e50dfddf157442a193c59b4699ebe562_2_1">
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<input type="radio" name="input_e50dfddf157442a193c59b4699ebe562_2_1" id="input_e50dfddf157442a193c59b4699ebe562_2_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="e50dfddf157442a193c59b4699ebe562_2_1-choice_0-label" for="input_e50dfddf157442a193c59b4699ebe562_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_e50dfddf157442a193c59b4699ebe562_2_1">
<span style="font-family: 'book antiqua', palatino;">True</span>
</label>
</div>
<div class="field">
<input type="radio" name="input_e50dfddf157442a193c59b4699ebe562_2_1" id="input_e50dfddf157442a193c59b4699ebe562_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="e50dfddf157442a193c59b4699ebe562_2_1-choice_1-label" for="input_e50dfddf157442a193c59b4699ebe562_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_e50dfddf157442a193c59b4699ebe562_2_1">
<span style="font-family: 'book antiqua', palatino;">False</span>
</label>
</div>
<span id="answer_e50dfddf157442a193c59b4699ebe562_2_1"/>
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<span class="status unanswered" id="status_e50dfddf157442a193c59b4699ebe562_2_1" data-tooltip="Not yet answered.">
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<div class="solution-span">
<span id="solution_e50dfddf157442a193c59b4699ebe562_solution_1"/>
</div><blockquote>
<p>
<span style="font-family: 'book antiqua', palatino;">No real number in <span class="math inline">\([0,1)\)</span> belongs to more than one <span class="math inline">\(V_q\)</span> (<span class="math inline">\(q \in \mathbb{Q}^{[0,1)}\)</span>).</span>
</p>
</blockquote>
<p>
<span style="font-family: 'book antiqua', palatino;">True or false?</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_e50dfddf157442a193c59b4699ebe562_3_1">
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<input type="radio" name="input_e50dfddf157442a193c59b4699ebe562_3_1" id="input_e50dfddf157442a193c59b4699ebe562_3_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="e50dfddf157442a193c59b4699ebe562_3_1-choice_0-label" for="input_e50dfddf157442a193c59b4699ebe562_3_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_e50dfddf157442a193c59b4699ebe562_3_1">
<span style="font-family: 'book antiqua', palatino;">True</span>
</label>
</div>
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<span style="font-family: 'book antiqua', palatino;">False</span>
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Problem 5
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<span style="font-family: 'book antiqua', palatino;">In showing that the Vitali Sets all have the same measure if they have a measure at all, we proceeded somewhat informally, by thinking of <span class="math inline">\([0,1)\)</span> as a circle. When <span class="math inline">\([0,1)\)</span> is instead thought of as a line-segment, one can get from <span class="math inline">\(V_0\)</span> to <span class="math inline">\(V_q\)</span> by translating <span class="math inline">\(V_0\)</span> by <span class="math inline">\(q\)</span>, and then subtracting 1 from any points that end up outside <span class="math inline">\([0,1)\)</span>. More precisely, <span class="math inline">\(V_0\)</span> can be transformed into <span class="math inline">\(V_q\)</span> in three steps. One first divides the points in <span class="math inline">\(V_0\)</span> into two subsets, depending on whether they are smaller than <span class="math inline">\(1-q\)</span>:</span>
</p>
<span style="font-family: 'book antiqua', palatino;">
<ul>
<li>
<p>
<span style="font-family: 'book antiqua', palatino;">
<span class="math inline">\(V_{\overset{\leftarrow}{0}} = V_0 \cap [0,1-q)\)</span>
</span>
</p>
</li>
<li>
<p>
<span style="font-family: 'book antiqua', palatino;">
<span class="math inline">\(V_{\overset{\rightarrow}{0}} = V_0 \cap [1-q,1)\)</span>
</span>
</p>
</li>
</ul>
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<p>
<span style="font-family: 'book antiqua', palatino;">Next, one translates <span class="math inline">\(V_{\overset{\leftarrow}{0}}\)</span> by <span class="math inline">\(q\)</span> and <span class="math inline">\(V_{\overset{\leftarrow}{0}}\)</span> by <span class="math inline">\(q-1\)</span>, yielding <span class="math inline">\((V_{\overset{\leftarrow}{0}})^q\)</span> and <span class="math inline">\((V_{\overset{\rightarrow}{0}})^{q-1}\)</span>, respectively. Finally, one takes the union of the translated sets: <span class="math inline">\((V_{\overset{\leftarrow}{0}})^q \cup (V_{\overset{\rightarrow}{0}})^{q-1}\)</span>.</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">Consider the following claim:</span>
</p>
<blockquote>
<p>
<span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(V_q = (V_{\overset{\leftarrow}{0}})^q \cup (V_{\overset{\rightarrow}{0}})^{q-1}\)</span> (<span class="math inline">\(q \in \mathbb{Q}^{[0,1)}\)</span>).</span>
</p>
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<p>
<span style="font-family: 'book antiqua', palatino;">True or false?</span>
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<h2 class="hd hd-2 unit-title">What to do about non-measurability?</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> In proving Vitali’s non-measurability theorem, we have learned a horrifying truth. We have learned that there is a precise sense in which it is impossible to assign a respectable measure to certain subsets of [mathjaxinline][0,1][/mathjaxinline]. How should one make sense of such a result? In this lecture we will explore some possible answers.</span></p>
<h4 class="p1"><span style="font-family: 'book antiqua', palatino;"><span class="Apple-converted-space"> </span>An Experiment to the Rescue?</span></h4>
<p><span style="font-family: 'book antiqua', palatino;">The intimate connection between measurability and probability raises an intriguing possibility. What if there was an “experiment” that allowed us to assign a measure to <span class="math inline">\(V\)</span>? Imagine a situation in which we are able to apply the Coin Toss Procedure again and again, to select points from <span class="math inline">\([0,1]\)</span>. Couldn’t we count the number of times we get a point in <span class="math inline">\(V\)</span>, and use this information to assign a measure to <span class="math inline">\(V\)</span>?</span></p>
<p><span style="font-family: 'book antiqua', palatino;">It goes without saying that it is impossible to carry out such an experiment in practice. Each application of the Coin Toss Procedure requires an infinite sequence of coin tosses. But let us idealize, and suppose that we are able to apply the Coin Toss Procedure (and indeed apply it infinitely many times). Notice, moreover, that in proving the theorem we didn’t actually lay down a definite criterion for membership in <span class="math inline">\(V\)</span>. All we did was show that some suitable set <span class="math inline">\(V\)</span> must exist. But let us idealize once more, and imagine that we are able to fix upon a specific Vitali set <span class="math inline">\(V\)</span>, by bypassing the Axiom of Choice and selecting a particular member from each family of “siblings” in <span class="math inline">\([0,1]\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Would such an idealized experiment help us assign a measure to <span class="math inline">\(V\)</span>? It seems to me that it would not. The first thing to note is that it is not enough to count the number of times the Coin Toss Procedure yields points <span class="math inline">\(V\)</span> and compare it with the number of times it yields points outside <span class="math inline">\(V\)</span>. To see this, imagine applying the Coin Toss Procedure <span class="math inline">\(\aleph_0\)</span>-many times, and getting the following results:</span></p>
<ul>
<li>
<p><span style="font-family: 'book antiqua', palatino;">Number of times procedure yielded a number in <span class="math inline">\(V\)</span>: <span class="math inline">\(\aleph_0\)</span>.</span></p>
</li>
<li>
<p><span style="font-family: 'book antiqua', palatino;">Number of times procedure yielded a number outside <span class="math inline">\(V\)</span>: <span class="math inline">\(\aleph_0\)</span>.</span></p>
</li>
</ul>
<p><span style="font-family: 'book antiqua', palatino;">Should you conclude that <span class="math inline">\(P(V) = 50\%\)</span>? No! Notice, for example, that in such an experiment one would expect to get <span class="math inline">\(\aleph_0\)</span>-many numbers in <span class="math inline">\([0,\frac{1}{4}]\)</span> and <span class="math inline">\(\aleph_0\)</span>-many numbers outside <span class="math inline">\([0,\frac{1}{4}]\)</span>. But that is no reason to think that <span class="math inline">\(p([0,\frac{1}{4}])=p\left(\overline{[0,\frac{1}{4}]}\right)=50\%\)</span>. If the experiment is to be helpful, we need more than just information about the <em>cardinality</em> of outputs within <span class="math inline">\(V\)</span> and outputs outside <span class="math inline">\(V\)</span>. We need a method for determining the <em>proportion</em> of total outputs that fall within <span class="math inline">\(V\)</span>. And, as we have seen, cardinalities need not determine proportions.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">One might wonder whether there is a strategy that one could use to try to identify the needed proportions “experimentally”. Suppose one carries out the Coin Toss Procedure infinitely many times, once for each natural number. One then determines, for each finite <span class="math inline">\(n\)</span>, how many of the first <span class="math inline">\(n\)</span> tosses fell within a given subset <span class="math inline">\(A\)</span> of <span class="math inline">\([0,1]\)</span>. (Call this number <span class="math inline">\(|A_n|\)</span>.) Finally, one lets the measure of <span class="math inline">\(A\)</span> be calculated experimentally, as follows: <span class="math display">\[\mu(A) = \lim_{n \to \infty} \frac{|A_n|}{n}\]</span> The hope is then that if one actually ran the experiment one would get the result that <span class="math inline">\(\mu([0,\frac{1}{4}]) = \frac{1}{4}\)</span>. More generally, one might hope to get the result that <span class="math inline">\(\mu(A)\)</span> turns out to be the Lebesgue Measure of <span class="math inline">\(A\)</span> whenever <span class="math inline">\(A\)</span> is a Borel Set (or, indeed, a Lebesgue Measurable set).</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let us suppose that we run such an experiment in an effort to identify a measure for our Vitali set <span class="math inline">\(V\)</span>. What sort of results one might expect? One possibility is that <span class="math inline">\(\mu(V)\)</span> fails to converge. In that case the experiment would give us no reason to think that <span class="math inline">\(V\)</span> has a measure. But suppose we run the experiment again and again, and that <span class="math inline">\(\mu(V)\)</span> always converges to the same number <span class="math inline">\(m \in [0,1]\)</span>. Would this give us any grounds for thinking that <span class="math inline">\(m\)</span> is the measure of <span class="math inline">\(V\)</span>? It is not clear that it would. For consider the following question: what values might our experiment assign to <span class="math inline">\(\mu(V_q)\)</span> (<span class="math inline">\(q \in \mathbb{Q}^{[0,1)}\)</span>)<span class="math inline"></span>?</span></p>
<p><span style="font-family: 'book antiqua', palatino;">There is no comfortable answer to this question. If we get the result that <span class="math inline">\(\mu(V_q) = m\)</span> for each <span class="math inline">\(q \in \mathbb{Q}^{[0,1)]}\)</span>, it will thereby be the case that <span class="math inline">\(\mu\)</span> fails to be countably additive, since there is no real number <span class="math inline">\(m\)</span> such that <span class="math inline">\(m+m+\dots = 1\)</span>. So we will have grounds for thinking that something has gone wrong with our experiment. If, on the other hand, we get the result that <span class="math inline">\(\mu(V_q)\)</span> is not uniformly <span class="math inline">\(m\)</span> for each <span class="math inline">\(q \in \mathbb{Q}^{[0,1)]}\)</span>, it will thereby be the case that <span class="math inline">\(\mu\)</span> fails to be uniform, since the \(V_q\) (\(q \in \mathbb{Q}^{[0,1)}\)) are essentially copies of \(V\), just positioned elsewhere in \([0,1)\). </span>So, again, we will have grounds for thinking that something has gone wrong with our experiment.</p>
<p><span style="font-family: 'book antiqua', palatino;">The moral is that it is not clear that we could use an experiment to help find a measure for <span class="math inline">\(V\)</span>, even when we take certain idealizations for granted. For should we find that <span class="math inline">\(\mu(V)\)</span> converges to some value, it is not clear that we would be left with grounds for assigning a measure to <span class="math inline">\(V\)</span>, rather than grounds for thinking that something went wrong with the experiment. In retrospect, this should come as no surprise. The non-measurability of <span class="math inline">\(V\)</span> follows from three very basic features of Lebesgue Measure: Non-Negativity, Countable Additivity and Uniformity. So we know that no version of our experiment could deliver a value for <span class="math inline">\(V\)</span> without violating at least one of these assumptions. And the price of violating the assumptions is that it’s no longer clear that the experiment delivers something worth calling a Lebesgue Measure.</span></p>
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<h2 class="hd hd-2 unit-title">Is there room for revising our assumptions?</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> When we proved that [mathjaxinline]V[/mathjaxinline] is nonmeasurable, what we proved is that there is no way of extending the Lebesgue Measure function [mathjaxinline]\lambda [/mathjaxinline] to a function that is defined for [mathjaxinline]V[/mathjaxinline] while satisfying all three of Non-Negativity, Countable Additivity and Uniformity. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Why is it so important that these assumptions be satisfied? </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">What would be so wrong about leaving Lebesgue Measure behind and instead focusing on a notion of measure that gives up on one or more of Non-Negativity, Countable Additivity and Uniformity? </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">When it comes to Uniformity, I think there is a simple answer: it seems to me that giving up on Uniformity means <em>changing the subject.</em> </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">The whole point of our enterprise is to find a way of extending the notion of Lebesgue Measure without giving up on uniformity. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">What about Non-Negativity and Countable Additivity? As it turns out, abandoning these constraints won’t help us avoid the phenomenon of non-measurability. When we discuss the Banach-Tarski Theorem in the next lecture we will see that there are sets that can be shown to be non-measurable without presupposing Non-Negativity or Countable Additivity.</span></p>
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<h2 class="hd hd-2 unit-title">Could we give up on the Axiom of Choice?</h2>
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<p><span style="font-family: 'book antiqua', palatino;">In order for <span class="math inline">\(V\)</span> to count as an example of a non-measurable set, it has to exist. And, as I mentioned earlier, <span class="math inline">\(V\)</span> cannot be shown to exist without the Axion of Choice:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Axiom of Choice</strong></span><br /><span style="font-family: 'book antiqua', palatino;">Every set of non-empty, non-overlapping sets has a choice set.</span></p>
<p></p>
<p><span style="font-family: 'book antiqua', palatino;">Because of this, one might be tempted to sidestep the phenomenon of non-measurability altogether, by giving up on the Axiom of Choice. Such a temptation might seem especially strong in light of the fact that the Axiom of Choice can be used to prove all sorts of bizarre results. We used it in Lecture 3 to find an incredible solution to a Hard Hat Problem, and we’ll soon use it to prove one of the most perplexing, and famous, of all its consequences: the Banach-Tarski Theorem.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Unfortunately, there are regions of mathematics that would be seriously weakened without the Axiom of Choice. For example, there would be no way of proving that, for any two sets, either they have the same size or one is bigger than the other:</span></p>
<p><span class="math display" style="font-family: 'book antiqua', palatino;">\[|A| = |B| \mbox{ or } |A| < |B| \mbox{ or } |B| < |A|\]</span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is hard to know what to do about the Axiom of Choice. When you think about some of its bizarre consequences, you might be tempted to give it up. But when you’re trying to work within certain areas of mathematics, you feel like you can’t live without it. My own view is that we should learn to live with the Axiom of Choice. But there is no denying that it delivers strange results.</span></p>
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<h2 class="hd hd-2 unit-title">[Optional: A Lesson for the Hat Problem]</h2>
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<h3><span style="font-family: 'book antiqua', palatino;">A Lesson for the Hat Puzzle</span></h3>
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<p><span style="font-family: 'book antiqua', palatino;">The existence of the Vitali set shows something interesting about probability — something that might ease our perplexity in the face of <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/5992963aeb8d44c393367c3ee7c22e9b" target="[object Object]">Bacon's Puzzle</a>, from Topic 6. See the video below to learn more.</span></p>
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<h3 class="hd hd-2">Video: A Lesson for the Hat Puzzle</h3>
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