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**Definition 89**.: Let \(A\) be a subset of \(\mathbb{R}\). The _outer measure_ of \(A\), denoted by \(m_{*}(A)\), is defined as
\[m_{*}(A)=\inf\left\{\sum_{k=1}^{\infty}|I_{k}|:\ \{I_{k}\}\text{ sequence of open intervals such that }\,A\operatorname{\subset}\bigcup_{k=1}^{\infty}I_{k}\right\},\]
where the infimum is taken over all countable coverings of \(A\).
The outer measure is always non-negative but could be infinite, so that in general we have \(0\leq m_{*}(A)\leq\infty\). Furthermore from the properties of infimum, for every \(\epsilon>0\), there exists a covering \(A\subset\bigcup_{k=1}^{\infty}I_{k}\) with
\[\sum_{k=1}^{\infty}|I_{k}|\leq m_{*}(A)+\epsilon.\]
##### Properties of the Outer Measure:
**Theorem 110**.:
* _Every countable subset of_ \(\mathbb{R}\) _has outer measure zero._
* \(m_{*}(A)=|A|\) _if_ \(A\) _is a closed interval._
* \(A\subset B\) _implies_ \(m_{*}(A)\leq m_{*}(B)\) _(Monotonicity)._
* _Suppose_ \(t\in\mathbb{R}\) _and_ \(A\subset\mathbb{R}\)_. Then_ \(m_{*}(A)=m_{*}(A+t)\) _(Translation invariance)._
* _If_ \(A=\bigcup_{j=1}^{\infty}A_{j}\)_, then_ \(m_{*}(A)\leq\sum_{j=1}^{\infty}m_{*}(A_{j})\) _(Countable sub-additivity)._
Proof
: For part a). Suppose \(A\) is countably many points and let \(A=\{x_{1},x_{2},\dots\}\) be an enumeration of the points of \(A\). Given \(\epsilon>0\), set \(I_{k}\) to be an open interval that contains the point \(x_{k}\) with length \(\frac{\epsilon}{2^{k}}\). The collection \(\{I_{k}\}\) covers \(A\) and \(\sum|I_{k}|=\sum\frac{\epsilon}{2^{k}}=\epsilon\); thus \(m_{*}(A)\leq\epsilon\).
For part b). Suppose \(A=[a,b]\). Given \(\epsilon>0\), one can set \(I_{1}=(a-\epsilon/2,\ b+\epsilon/2)\)and for \(k>1\) set \(I_{k}=\emptyset.\) Thus, \(m_{*}(A)\leq b-a+\epsilon\). Because this inequality holds for every \(\epsilon>0\), we conclude that \(m^{*}(A)\leq b-a\). For the converse inequality suppose \(\{I_{k}\}\) to be an open cover for \(A\). Since \(A\) is compact, there is a finite subcover, \(A\subset\bigcup_{j=1}^{N}I_{j}\). Renumbering we get points
\[a=a_{1}<\cdots Then \[|A|=b-a=a_{n+1}-a_{1}\leq\sum_{j=1}^{n}|I_{j}|\ \ \ \mbox{thus}\ \ \ |A|\leq m_{*}(A).\] Monotonicity and translation invariance properties follows from the definition of outer measure. For part e) (Countable sub-additivity). We may assume that \(m_{*}(A_{j})<\infty\). If \(A=\bigcup_{j=1}^{\infty}A_{j}\), then for any \(\epsilon>0\), the definition of outer measure yields for each \(j\) a covering \(A_{j}\subset\bigcup_{k=1}^{\infty}I_{k,j}\) with \[\sum_{k=1}^{\infty}|I_{k,j}|\leq m_{*}(A_{j})+\frac{\epsilon}{2^{j}}.\] Thus, \(A\subset\bigcup_{j,k=1}^{\infty}I_{k,j}\) is a covering for \(A\), and therefore \[m_{*}(A)\leq\sum_{j,k}|I_{k,j}|=\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}|I_{k,j} |\leq\sum_{j=1}^{\infty}\left(m_{*}(A_{j})+\frac{\epsilon}{2^{j}}\right)=\sum _{j=1}^{\infty}m_{*}(A)+\epsilon\] holds for every \(\epsilon>0\); whence the countable sub-additivity is proved. **Corollary 21**.: _Every interval \(B\) in \(\mathbb{R}\) that contains at least two distinct points \(a,b\in\mathbb{R}\) with \(a
: The monotonicity property of outer measure implies \(|B|\geq|[a,b]|=b-a\) and because every countable subset of \(\mathbb{R}\) has outer measure \(0\) (see property a)), we can conclude that \(B\) is uncountable. Suppose \(A\) and \(B\) are subsets of \(\mathbb{R}\) such that \(A\cap B=\emptyset\). Do we have \[m_{*}(A\cup B)=m_{*}(A)+m_{*}(B)?\] This important property is not in the above list of properties of the outer measure for a good reason. It holds when the sets are not “irregular” but are measurable. Rather than modify \(m_{*}\), we restrict ourselves to the class of “measurable” subsets that we work with. ### 2.2 Real Analysis #### 2.2.1 The \(\mathbb{R}\)-matrix The \(\mathbb{R}\)-matrix is defined as \[\mathbb{R}=\left\{\begin{array}{ll}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{array}\right.\] where \(\mathbb{R}\) is the set of all possible values of \(\mathbb{R}\). #### 2.2.2 The \(\mathbb{R}\)-matrix The \(\mathbb{R}\)-matrix is defined as \[\mathbb{R}=\left\{\begin{array}{ll}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{array}\right. * (Approximation). Suppose \(A\) is a measurable subset of \(\mathbb{R}\). Then for every \(\epsilon>0\) there exists a closed set \(C\subset A\) with \[m(A\setminus C)\leq\epsilon.\] If \(m(A)\) is finite, there exists a compact set \(K\) with \(K\subset A\) and \[m(A\setminus K)\leq\epsilon.\] **Remark 60**.: Are all subsets of \(\mathbb{R}\) measurable? The answer to this question is given in the section on the Banach-Tarski paradox (see Section 2.8), where we present a construction of a subset of \(\mathbb{R}\) which is not measurable. ### Measurable Functions If \(f:\mathbb{R}\to\mathbb{R}\), recall that \(f\) is continuous if and only if \(f^{-1}(A)\) is open whenever \(A\) is open. Here the inverse image \(f^{-1}(A)\) of \(A\) is defined as \[f^{-1}(A)=\left\{x\in\mathbb{R}:f(x)\in A\right\}.\] In this section we will define measurable functions which are “almost” continuous, but in a somewhat weaker sense than that of Riemann integrable functions. That is what we expect, since we want the class of Lebesgue integrable functions to be larger than the class of Riemann integrable functions. Note that we say \(f\) is finite valued if \(-\infty **Definition 91**.: A real-valued function \(f\) defined on \(\mathbb{R}\) is a _measurable function_ if for each open set \(A\) of \(\mathbb{R}\), the inverse image \(f^{-1}(A)\) is measurable. Note that this property also applies to extended-valued functions if we make the additional hypothesis that both \(f^{-1}(\infty)\) and \(f^{-1}(-\infty)\) are measurable sets. Also if \(f\) is a continuous function with a measurable domain then \(f\) is measurable, since open sets are measurable. Moreover monotone functions, step functions, and semicontinuous functions defined on some interval in \(\mathbb{R}\) are all measurable. In particular if we think of \(A=[-\infty,a)\), \[f^{-1}([-\infty,a))=\left\{x\in\mathbb{R}:\ f
and write \(\left\{f
* \(\left\{x:f(x)>a\right\}\) is measurable for all \(a\in\mathbb{R}\). * \(\{x:f(x)\leq a\}\) is measurable for all \(a\in\mathbb{R}\). Using set relations together with properties of measurable sets will show the above equivalencies. For example, \(a)\) implies \(b)\), rewriting the set \(\{f\geq a\}\) as \[\{f\geq a\}=f^{-1}([a,\infty))=f^{-1}\left(\bigcap_{k=1}^{\infty}(a-1/k,\infty)\right)\] and recalling that the inverse image from sets to sets preserves countable intersections (actually arbitrary intersection, unions, and complements) we have: \[f^{-1}\left(\bigcap_{k=1}^{\infty}(a-1/k,\infty)\right)=\bigcap_{k=1}^{\infty }f^{-1}\left((a-1/k,\infty)\right)=\bigcap_{k=1}^{\infty}\{f>a-1/k\}\] and the intersection of a sequence of measurable sets is measurable. The following proposition tells us that certain operations performed on measurable functions yield again measurable functions. **Definition 92**.: If \(f\) and \(g\) are two real-valued functions on \(\mathbb{R}\), then we define \[(f\wedge g)(x)=\min\{f(x),g(x)\},\qquad(f\lor g)(x)=\max\{f(x),g(x)\}.\]
Proof
* \(\left\{x:f(x)\geq a\right\}\) is measurable for all \(a\in\mathbb{R}\).
* \(\left\{x:f(x)
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