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<h2 class="hd hd-2 unit-title">Measure</h2>
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<p><span style="font-family: 'book antiqua', palatino;">The easiest way to get clear about the notion of measure is to start by thinking about the notion of <em>length</em>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let us start by considering the notion of length as it applies to line segments <span class="math inline">\([a,b]\)</span>, where <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> are real numbers such that <span class="math inline">\(a \leq b\)</span>. Although it is natural to think of of line segments in spatial terms, here we will model them as sets of real numbers. More specifically, we will think of the line segment <span class="math inline">\([a,b]\)</span> as the set of real numbers <span class="math inline">\(x\)</span> such that <span class="math inline">\(a \leq x \leq b\)</span>. So, for instance, we take <span class="math inline">\([\frac{1}{4},\frac{1}{2}]\)</span> to be the set <span class="math inline">\(\left\{x:\frac{1}{4} \leq x \leq \frac{1}{2}\right\}\)</span>. (We will treat <span class="math inline">\([a,a] = \{a\}\)</span> as a line segment, and therefore regard a point as a special case of a line segment.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">A nice feature of line segments is that there is a simple recipe for characterizing their length:</span></p>
<blockquote>
<p><span style="font-family: 'book antiqua', palatino;">Length<span class="math inline">\(\left([a,b]\right) = b - a\)</span>.</span></p>
</blockquote>
<p><span style="font-family: 'book antiqua', palatino;">Accordingly, <span class="math inline">\([\frac{1}{2},\frac{3}{4}]\)</span> has length <span class="math inline">\(\frac{1}{4}\)</span>, because <span class="math inline">\(\frac{3}{4} - \frac{1}{2} = \frac{1}{4}\)</span>; and <span class="math inline">\([2,2]\)</span> has length <span class="math inline">\(0\)</span>, because <span class="math inline">\(2-2 = 0\)</span>. Now suppose that we wish to generalize the notion of length by applying it to sets of real numbers other than line segments. Suppose, for example, that we wish to assign a “length” to the set: <span class="math display">\[\left[0,\frac{1}{4}\right] \cup \left[\frac{1}{2},1\right]\]</span></span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is natural to say the following:</span></p>
<p><span class="math display" style="font-family: 'book antiqua', palatino;">\[\text{Length}\left(\left[0,\frac{1}{4}\right] \cup \left[\frac{1}{2},1\right]\right) = \text{Length}\left([0,\frac{1}{4}]\right) + \text{Length}\left([\frac{1}{2},1]\right) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\]</span></p>
<p><span style="font-family: 'book antiqua', palatino;">And as it turns out, there is a systematic way of extending this general idea to an rich and interesting family of sets of real numbers. They are known as the <em>Borel Sets</em>, in honor of the great French mathematician Émile Borel. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">I’ll tell you more about them in the next subsection.</span></p>
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<h2 class="hd hd-2 unit-title">The Borel Sets</h2>
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<p><span style="font-family: 'book antiqua', palatino;">A <strong>Borel Set</strong> is a set that you can get to by starting with the family of line segments and carrying out finitely many applications of the operations of <em>complementation</em> and <em>countable union</em>:</span></p>
<ul>
<li>
<p><span style="font-family: 'book antiqua', palatino;">The <strong>complementation operation</strong> takes each set <span class="math inline">\(A\)</span> to <span class="math inline">\(\mathbb{R}-A\)</span>, where <span class="math inline">\(\mathbb{R}\)</span> is the set of real numbers. (I will sometimes refer to the complement of <span class="math inline">\(A\)</span> as <span class="math inline">\(\overline{A}\)</span>.) So, for instance, the result of applying the complement operation to <span class="math inline">\([0,1]\)</span> is the set <span class="math inline">\(\overline{[0,1]} = \mathbb{R}-[0,1]\)</span>, which consists of every real number except for the elements of <span class="math inline">\([0,1]\)</span>.</span></p>
</li>
<li>
<p><span style="font-family: 'book antiqua', palatino;">The <strong>countable union operation</strong> takes each countable (i.e. finite or countably infinite) family of sets <span class="math inline">\(A_1, A_2, A_3, \ldots\)</span> to their union: <span class="math inline">\(\bigcup \{A_1, A_2 , A_3 \dots\}\)</span>. So, for instance, the result of applying the countable union operation to the sets <span class="math inline">\([0,\frac{1}{2}], [\frac{1}{2},\frac{3}{4}], [\frac{3}{4},\frac{7}{8}],\dots\)</span> is the set <span class="math inline">\([0,1)\)</span>, which consists of the real numbers <span class="math inline">\(x\)</span> such that <span class="math inline">\(0 \leq x < 1\)</span>. (Note that the round bracket on the right-hand side of “<span class="math inline">\([0,1)\)</span>” is used to indicate that the end-point <span class="math inline">\(1\)</span> is not included in the set.)</span></p>
</li>
</ul>
<p><span style="font-family: 'book antiqua', palatino;">Formally, a Borel Set is any member of the smallest class <span class="math inline">\(\mathscr{B}\)</span> such that: <span class="math inline">\((i)\)</span> every line-segment is in <span class="math inline">\(\mathscr{B}\)</span>, <span class="math inline">\((ii)\)</span> if a set is in <span class="math inline">\(\mathscr{B}\)</span>, then so is its complement, and <span class="math inline">\((iii)\)</span> if a countable family of sets is in <span class="math inline">\(\mathscr{B}\)</span>, then so is its union.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">To get some practice working with Borel Sets, let us verify that the set of <em>irrational</em> real numbers <span class="math inline">\(\overline{\mathbb{Q}}\)</span> is a Borel Set. (<span class="math inline">\(\overline{\mathbb{Q}}=\mathbb{R} - \mathbb{Q}\)</span>, where <span class="math inline">\(\mathbb{Q}\)</span> is the set of rational numbers.) What we need to show is that the set <span class="math inline">\(\overline{\mathbb{Q}}\)</span> can be generated by starting with a family of line segments <span class="math inline">\([a,b]\)</span> and applying complementation and countable union as many times as needed. This can be done in three steps:</span></p>
<ul>
<li>
<p><span style="font-family: 'book antiqua', palatino;"><em>Step 1</em>: For every rational number <span class="math inline">\(q \in \mathbb{Q}\)</span>, the singleton set <span class="math inline">\(\{q\}\)</span> is identical to the point-sized line-segment <span class="math inline">\([q,q]\)</span>. Since every line-segment is a Borel Set, it follows that every singleton set <span class="math inline">\(\{q\}\)</span> (<span class="math inline">\(q \in \mathbb{Q}\)</span>) is a Borel Set.</span></p>
</li>
<li>
<p><span style="font-family: 'book antiqua', palatino;"><em>Step 2</em>: Since the set of rational numbers, <span class="math inline">\(\mathbb{Q}\)</span>, is countable, it is the countable union of the family of Borel Sets <span class="math inline">\(\{q\}\)</span>, for <span class="math inline">\(q \in \mathbb{Q}\)</span>. And since the countable union of Borel Sets is a Borel Set, <span class="math inline">\(\mathbb{Q}\)</span> must be a Borel Set too.</span></p>
</li>
<li>
<p><span style="font-family: 'book antiqua', palatino;"><em>Step 3</em>: Since <span class="math inline">\(\mathbb{Q}\)</span> is a Borel Set, <span class="math inline">\(\overline{\mathbb{Q}}\)</span> is the complement of a Borel Set. So <span class="math inline">\(\overline{\mathbb{Q}}\)</span> must be a Borel Set too.</span></p>
</li>
</ul>
<p><span style="font-family: 'book antiqua', palatino;">(Note that a procedure of this kind can be used to show that any countable set is a Borel Set, as is its complement.)</span></p>
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<span style="font-family: 'book antiqua', palatino;">If <span class="math inline">\(A_1, A_2, A_3, \dots\)</span> is a countable family of Borel Sets, then <span class="math inline">\(\bigcap\{A_1, A_2, A_3, \dots\}\)</span> is a Borel Set.</span>
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<p>
<span style="font-family: 'book antiqua', palatino;">(In general, the <strong>intersection</strong> of a set <span class="math inline">\(\{A_1, A_2, A_3, \dots\}\)</span> (in symbols: <span class="math inline">\(\bigcap\{A_1, A_2, A_3, \dots\}\)</span>) is the set of individuals <span class="math inline">\(x\)</span> such that <span class="math inline">\(x\)</span> is a member of each set <span class="math inline">\(A_1, A_2, A_3, \dots\)</span>.)</span>
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<span style="font-family: 'book antiqua', palatino;">True or false?</span>
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<span style="font-family: 'book antiqua', palatino;">Assume that <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are Borel Sets. Is it true that <span class="math inline">\(A - B\)</span> (i.e. the set of elements in <span class="math inline">\(A\)</span> which are not in <span class="math inline">\(B\)</span>) is a Borel Set?</span>
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<h2 class="hd hd-2 unit-title">Lebesgue Measure</h2>
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<p><span style="font-family: 'book antiqua', palatino;">As it turns out, there is <em>exactly one</em> natural way of extending the ordinary notion of length so that it applies to all Borel Sets. More precisely, there is exactly one function <span class="math inline">\(\lambda\)</span> on the Borel Sets that satisfies these three conditions:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Length on Segments</strong></span><br /><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(\lambda([a,b]) = b-a\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">(This condition is meant to ensure that <span class="math inline">\(\lambda\)</span> counts as an extension, rather than a modification, of the notion of length.)</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Countable Additivity</strong></span><br /><span style="font-family: 'book antiqua', palatino;">Let <span class="math inline">\(A_1,A_2,\dots\)</span> be a countable family of disjoint sets. Whenever <span class="math inline">\(\lambda(A_i)\)</span> is defined for each <span class="math inline">\(A_i\)</span>, we have: <span class="math display">\[\lambda\left(\bigcup\{A_1, A_2 , A_3,\ldots\}\right) = \lambda(A_1) + \lambda(A_2) + \lambda(A_3) + \ldots\]</span> (For <span class="math inline">\(A_1,A_2,\dots\)</span> to be <em>disjoint</em> is for <span class="math inline">\(A_i\)</span> and <span class="math inline">\(A_j\)</span> to have no elements in common whenever <span class="math inline">\(i \neq j\)</span>.)</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Non-Negativity</strong></span><br /><span style="font-family: 'book antiqua', palatino;">For any set <span class="math inline">\(A\)</span> in the domain of <span class="math inline">\(\lambda\)</span>, <span class="math inline">\(\lambda(A)\)</span> is either a non-negative real number, or the infinite value <span class="math inline">\(\infty\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">(Note that the length of a line-segment <span class="math inline">\([a,b]\)</span> is always a non-negative real number. When we transition from measuring line-segments to measuring Borel Sets, however, we allow for sets of “infinite length”, such as <span class="math inline">\([0,\infty)\)</span>, which is the set of non-negative real numbers, or <span class="math inline">\((-\infty,\infty) = \mathbb{R}\)</span>.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The unique function <span class="math inline">\(\lambda\)</span> on the Borel Sets that satisfies these three conditions is called the <strong>Lebesgue Measure</strong>, in honor of another great French mathematician: Henry Lebesgue.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">We will not prove that <span class="math inline">\(\lambda\)</span> exists here, or that it is unique. We will simply assume that <span class="math inline">\(\lambda\)</span> exists and is well-defined for every Borel Set. (If you’d like to learn how to prove the relevant result, I recommend a measure-theory textbook in Lecture 7.3.)</span></p>
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Problem 6
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<span style="font-family: 'book antiqua', palatino;">Countable Additivity gives us an additivity condition for finite or countably infinite families of disjoint sets. Would it be a good idea to insist on an additivity condition for <em>uncountable</em> families of disjoints sets?</span>
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<h2 class="hd hd-2 unit-title">[Optional: Extending Lebesgue Measure Beyond the Borel Sets]</h2>
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<p><span style="font-family: 'book antiqua', palatino;">As it turns out, there is a natural way of extending the notion of Lebesgue Measure beyond the Borel Sets while preserving the crucial properties of Countable Additivity and Non-Negativity. (The property of Length on Segments is preserved automatically by any extension of <span class="math inline">\(\lambda\)</span>.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The key idea is that any subset of a Borel Set of measure zero should be treated as having measure zero. Formally speaking, one extends the notion of Lebesgue Measure in two steps. The first step is to introduce a notion of “Lebesgue measurability”. We say that <span class="math inline">\(A\)</span> is <strong>Lebesgue Measurable</strong> if and only if:</span></p>
<ol>
<li>
<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(A\)</span> is a Borel Set; or</span></p>
</li>
<li>
<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(A = A^B \cup A^0\)</span> where <span class="math inline">\(A^B\)</span> is a Borel Set and <span class="math inline">\(A^0\)</span> is a subset of some Borel Set of Lebesgue Measure zero.</span></p>
</li>
</ol>
<p><span style="font-family: 'book antiqua', palatino;">The second step is to extend the function <span class="math inline">\(\lambda\)</span> by stipulating that <span class="math inline">\(\lambda(A) = \lambda(A^B)\)</span>, whenever <span class="math inline">\(A\)</span> and <span class="math inline">\(A^B\)</span> are as above.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">This procedure doesn’t affect the value of <span class="math inline">\(\lambda\)</span> on Borel Sets. And, as you’ll verify in the exercises below, it yields the result that Countable Additivity and Non-Negativity hold for any Lebesgue Measurable sets, whether or not they are Borel Sets.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is worth noting that our definition of Lebesgue measurability wouldn’t get us beyond the Borel Sets if the only Borel Sets with measure zero were countable sets, since every subset of a countable set is countable (and therefore a Borel Set), and since the union of two Borel Sets is always a Borel Set. As it turns out, however, there are <em>uncountable</em> Borel Sets of Lebesgue Measure zero which have subsets that are not Borel Sets. The most famous example of an uncountable set of Lebesgue Measure zero is the <em>Cantor Set</em>, which is named in honor of our old friend from Lecture 1, Georg Cantor. We won’t get into it here, but you can find plenty of information about it on the Web.</span></p>
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<span style="font-family: 'book antiqua', palatino;">Suppose that that <span class="math inline">\(A^0\)</span> is a subset of a Borel Set with Lebesgue Measure zero. Is it then true that <span class="math inline">\(\lambda(A^0) = 0\)</span>?</span>
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Problem 4
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<span style="font-family: 'book antiqua', palatino;">Whenever <span class="math inline">\(A\)</span> is Lebesgue Measurable, <span class="math inline">\(\lambda(A)\)</span> is either a non-negative real number or <span class="math inline">\(\infty\)</span>, and therefore Non-Negativity is preserved when <span class="math inline">\(\lambda\)</span> is extended to all Lebesgue Measurable sets.</span>
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Problem 5
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<span style="font-family: 'book antiqua', palatino;">Our extension of <span class="math inline">\(\lambda\)</span> to the Lebesgue Measurable sets preserves Countable Additivity.</span>
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<span style="font-family: 'book antiqua', palatino;">In other words: assume that <span class="math inline">\(A_1,A_2,\dots\)</span> are a finite or countably infinite family of disjoint sets. Whenever <span class="math inline">\(A_i\)</span> is Lebesgue Measurable for each <span class="math inline">\(i\)</span>, we have: <span class="math display">\[\lambda\left(A_1 \cup A_2 \cup A_3 \cup \ldots\right) = \lambda(A_1) + \lambda(A_2) + \lambda(A_3) + \ldots\]</span></span>
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<h2 class="hd hd-2 unit-title">Uniformity</h2>
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<p><span style="font-family: 'book antiqua', palatino;">The Lebesgue Measure is one kind of measure on the Borel Sets. But there are others. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">A function on the Borel Sets is said to be a <strong>measure</strong> if and only if it satisfies Countable Additivity and Non-Negativity (and for it to assign the value 0 to the empty set).</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Accordingly, the Lebesgue Measure is a measure with the additional property of satisfying Length on Segments. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">This is important because it makes the Lebesgue Measure <em>uniform</em>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Intuitively speaking, what it means for a measure to be uniform is for it to be such that the measure of a set always remains unchanged when the set is moved to a different location within the real line. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Formally speaking, a measure <span class="math inline">\(\mu\)</span> is uniform if and only if:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Uniformity</strong></span><br /><span style="font-family: 'book antiqua', palatino;">Let <span class="math inline">\(\mu(A)\)</span> be well-defined, and let <span class="math inline">\(A^c\)</span> be the result of translating <span class="math inline">\(A\)</span> by <span class="math inline">\(c \in \mathbb{R}\)</span>. Then <span class="math inline">\(\mu(A^c)\)</span> is well-defined, and equal to <span class="math inline">\(\mu(A)\)</span>.</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;">(The <strong>translation</strong> of <span class="math inline">\(A\)</span> by <span class="math inline">\(c\)</span> is the result of adding <span class="math inline">\(c\)</span> to each member of <span class="math inline">\(A\)</span>. In other words: <span class="math inline">\(A^c = \{x+c : x \in A\}\)</span>. Intuitively, you can think of the translation of <span class="math inline">\(A\)</span> by <span class="math inline">\(c\)</span> as the result of rigidly moving <span class="math inline">\(A\)</span> <span class="math inline">\(c\)</span> units to the right.)</span></p>
<p></p>
<p><span style="font-family: 'book antiqua', palatino;">Here is an example. The Lebesgue Measure of <span class="math inline">\([0,\frac{1}{4}]\)</span> is <span class="math inline">\(\frac{1}{4}\)</span>. So the uniformity of Lebesgue Measure entails that the Lebesgue Measure of <span class="math inline">\([0,\frac{1}{4}]^{\frac{1}{2}}\)</span> (which is the result of translating <span class="math inline">\([0,\frac{1}{4}]\)</span> by <span class="math inline">\(\frac{1}{2}\)</span>) must also equal <span class="math inline">\(\frac{1}{4}\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Uniformity is such a natural property that it is tempting to take it for granted. But there are all sorts of measures that do not satisfy Uniformity. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">We’ll talk about one of them next.</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">So far we have been thinking of the notion of measure as a generalization of the notion of <em>length</em>. But, as I suggested at the beginning of this chapter, the notion of measure can also be thought of as a generalization of the notion of <em>probability</em>. Suppose, for example, that we have a random procedure for selecting individual points from the line-segment <span class="math inline">\([0,1]\)</span>. For <span class="math inline">\(A\)</span> a subset of <span class="math inline">\([0,1]\)</span>, let <span class="math inline">\(p(A)\)</span> be the probability that the next point selected from <span class="math inline">\([0,1]\)</span> by our random procedure is a member of <span class="math inline">\(A\)</span>. As long as <span class="math inline">\(p(\ldots)\)</span> satisfies Countable Additivity, it will count as a measure on the Borel Sets in [0,1]. But whether or not <span class="math inline">\(p(\dots)\)</span> turns out to be the Lebesgue Measure will depend on the details of our random selection procedure. The best way to see this is to consider two different kinds of selection procedures:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Standard Coin-Toss Procedure</strong></span><br /><span style="font-family: 'book antiqua', palatino;">You toss a fair coin once for each natural number. Each time the coin lands Heads you write down a zero, and each time it lands Tails you write down a one. This gives you an infinite sequence <span class="math inline">\(\langle d_1,d_2,d_3,\ldots\rangle\)</span>, The selection procedure is then as follows: pick whichever number in <span class="math inline">\([0,1]\)</span> has <span class="math inline">\(0.d_1d_2d_3\ldots\)</span> as its binary expansion.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">(If you’d like a refresher on binary expansions, have a look at Lecture 1.5.1. As noted in that discussion, the rational numbers within [0,1] have two different binary expansions: one ending in 0s and the other ending in 1s. To simplify the present discussion, I assume that the output of the Coin-Toss Procedure is declared void if it corresponds to a binary expansion ending in 1s.)</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Square Root Coin-Toss Procedure</strong></span><br /><span style="font-family: 'book antiqua', palatino;">As before, but this time you pick the <em>square root</em> of the number represented by <span class="math inline">\(0.d_1d_2d_3\ldots\)</span>. (Suppose, for instance, that sequence of coin-tosses yields 0.0100(0), which is the binary expansion of <span class="math inline">\(\frac{1}{4}\)</span>. Then you should select <span class="math inline">\(\frac{1}{2}\)</span>, since <span class="math inline">\(\frac{1}{2} = \sqrt{\frac{1}{4}}\)</span>.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The Standard Coin-Toss Procedure delivers a uniform probability distribution. In fact, when one makes suitable assumptions about the probabilities of sequences of coin tosses, the Standard Coin Toss Procedure generates a Lebesgue Measure over <span class="math inline">\([0,1]\)</span>. The Square Root Coin Toss Procedure, in contrast, does not satisfy Uniformity. It can be used to define a measure over <span class="math inline">\([0,1]\)</span>, but not the Lebesgue Measure.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here is an intuitive way of visualizing the difference between the two measures. Suppose we have 1kg of mud, and that we are told to pour it on the line segment <span class="math inline">\([0,1]\)</span> in such a way that the amount of mud above an interval within <span class="math inline">\([0,1]\)</span> is proportional to the probability that the next point selected by one of our coin-toss procedures will fall within that interval. In the case of the Standard Coin-Toss Procedure, our mud distribution should look like this:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-14.png" type="saveimage" target="[object Object]" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">In other words: the probability of getting a number within a given interval does not depend on where the interval is located within <span class="math inline">\([0,1]\)</span>: it depends only on the size of the interval. When it comes to the Square Root Coin-Toss Procedure, in contrast, our mud distribution should look like this:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-14final.png" width="1088" height="365" type="saveimage" target="[object Object]" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">In other words: the probability of getting a number within a given interval depends not just on the size of the interval, but also on the <em>location</em> of the interval within <span class="math inline">\([0,1]\)</span>: the closer to 1 the interval is, the higher the probability of getting a number within that interval.</span></p>
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<p>
<span style="font-family: 'book antiqua', palatino;">Answer the following questions with respect to each of our two coin-toss procedures:</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">(1) What is the probability that the next selected point will be in <span class="math inline">\([\frac{1}{2},1]\)</span>?</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">Standard Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_2_1" class="inputtype formulaequationinput">
<div class="unanswered">
<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_2_1" id="input_0aa062278d62494b9d2be020f0cbb787_2_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_2_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_2_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_2_1"/>
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<div id="input_0aa062278d62494b9d2be020f0cbb787_2_1_preview" class="equation">
\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
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<p>
<span style="font-family: 'book antiqua', palatino;">Square Root Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_3_1" class="inputtype formulaequationinput">
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<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_3_1" id="input_0aa062278d62494b9d2be020f0cbb787_3_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_3_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_3_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_3_1"/>
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<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<span id="solution_0aa062278d62494b9d2be020f0cbb787_solution_1"/>
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<br/>
<p>
<span style="font-family: 'book antiqua', palatino;">(2) What is the probability that the next selected point will be precisely <span class="math inline">\(\frac{1}{2}\)</span>? (Assume that the relevant probability is defined and is a real number between 0 and 1.)</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">Standard Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_4_1" class="inputtype formulaequationinput">
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<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_4_1" id="input_0aa062278d62494b9d2be020f0cbb787_4_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_4_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_4_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_4_1"/>
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<div id="input_0aa062278d62494b9d2be020f0cbb787_4_1_preview" class="equation">
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<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
</div></div>
<p>
<span style="font-family: 'book antiqua', palatino;">Square Root Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 4" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_5_1" class="inputtype formulaequationinput">
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<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_5_1" id="input_0aa062278d62494b9d2be020f0cbb787_5_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_5_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_5_1" size="20"/>
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<span id="solution_0aa062278d62494b9d2be020f0cbb787_solution_2"/>
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<br/>
<p>
<span style="font-family: 'book antiqua', palatino;">(3) What is the probability that the next selected point will be in <span class="math inline">\([0,\frac{1}{2}]\)</span>? (<em>Hint:</em> use your answer to the previous question, and assume that the relevant probabilities satisfy Additivity.)</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">Standard Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 5" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_6_1" class="inputtype formulaequationinput">
<div class="unanswered">
<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_6_1" id="input_0aa062278d62494b9d2be020f0cbb787_6_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_6_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_6_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_6_1"/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<div id="input_0aa062278d62494b9d2be020f0cbb787_6_1_preview" class="equation">
\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
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<p>
<span style="font-family: 'book antiqua', palatino;">Square Root Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 6" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_7_1" class="inputtype formulaequationinput">
<div class="unanswered">
<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_7_1" id="input_0aa062278d62494b9d2be020f0cbb787_7_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_7_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_7_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_7_1"/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<div id="input_0aa062278d62494b9d2be020f0cbb787_7_1_preview" class="equation">
\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
</div>
</div>
<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
</div></div>
<div class="solution-span">
<span id="solution_0aa062278d62494b9d2be020f0cbb787_solution_3"/>
</div><p/>
<br/>
<p>
<span style="font-family: 'book antiqua', palatino;">(4) Suppose <span class="math inline">\(\{x_0, x_1, x_2, \dots\}\)</span> is a countable set of numbers in <span class="math inline">\([0,1]\)</span>. What is the probability that the next selected point will be in <span class="math inline">\(\{x_0, x_1, x_2, \dots\}\)</span>? (Assume that the relevant probabilities satisfy Countable Additivity.) <em>Hint:</em> Use your answer to the questions above.</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">Standard Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 7" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_8_1" class="inputtype formulaequationinput">
<div class="unanswered">
<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_8_1" id="input_0aa062278d62494b9d2be020f0cbb787_8_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_8_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_8_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_8_1"/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<p id="answer_0aa062278d62494b9d2be020f0cbb787_8_1" class="answer"/>
<div id="input_0aa062278d62494b9d2be020f0cbb787_8_1_preview" class="equation">
\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
</div></div>
<p>
<span style="font-family: 'book antiqua', palatino;">Square Root Coin-Toss Procedure:</span>
</p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 8" role="group"><div id="formulaequationinput_0aa062278d62494b9d2be020f0cbb787_9_1" class="inputtype formulaequationinput">
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<input type="text" name="input_0aa062278d62494b9d2be020f0cbb787_9_1" id="input_0aa062278d62494b9d2be020f0cbb787_9_1" data-input-id="0aa062278d62494b9d2be020f0cbb787_9_1" value="" aria-describedby="status_0aa062278d62494b9d2be020f0cbb787_9_1" size="20"/>
<span class="trailing_text" id="trailing_text_0aa062278d62494b9d2be020f0cbb787_9_1"/>
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<div id="input_0aa062278d62494b9d2be020f0cbb787_9_1_preview" class="equation">
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<span id="solution_0aa062278d62494b9d2be020f0cbb787_solution_4"/>
</div><p/>
<br/>
<p>
<span style="font-family: 'book antiqua', palatino;">(5) What is the probability that the next point selected will be a member of <span class="math inline">\([0,1]-\{\frac{1}{2}\}\)</span>? What is the probability that it will be a member of <span class="math inline">\([0,1]-\{x_0, x_1, x_2, \dots\}\)</span>? (Assume that the relevant probabilities are all well-defined, and that they satisfy Countable Additivity.)</span>
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<span style="font-family: 'book antiqua', palatino;">Standard Coin-Toss Procedure:</span>
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<span style="font-family: 'book antiqua', palatino;">Square Root Coin-Toss Procedure:</span>
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