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Q) Suppose that $f$ is a simple Lebesgue measurable function on $R^n$. Show that there exists a Borel measurable function $g$ such that $f(x) = g(x)$ for almost every $x \in R^n$.

I know that a simple Lebesgue measurable function takes finitely many values on Lebesgue measurable sets. But I'm not sure how I can find an appropriate Borel measurable function $g$, i.e. function wherein the pre-image of every open set is Borel measurable, which is the same as $f$ except atmost on a countable set?

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The problem boils down to: For each Lebesgue measurable set $A$, there exists a Borel set $B$ such that $m(A\Delta B)=0$. For simplicity, I demonstrate the proof for the case $n=1$.

Proof: Let $A$ be Lebesgue measurable. Firstly, consider the case that $m(A)<\infty$. For each $n$, there exists an open set $U_{n}$ such that $A\subseteq U_{n}$ and $m(U_{n})-m(A)<\frac{1}{n}$. Define $B=\cap_{n}U_{n}$, which is a Borel set (Actually, we can say more: it is a $G_{\delta}$-set. However, we do not need this fact.) Clealry $A\subseteq B$. Moreover $B\setminus A\subseteq U_{n}\setminus A$ implies that $m(B\setminus A)\leq m(U_{n}\setminus A)=m(U_{n})-m(A)<\frac{1}{n}$. Since $n$ is arbitrary, we have $m(B\setminus A)=0$. Next, we drop the assumption that $m(A)<\infty$. For each $n\in\mathbb{Z}$, let $A_{n}=A\cap(n,n+1]$. For each $n$, choose a Borel set $B_{n}$ such that $A_{n}\subseteq B_{n}$ and $m(B_{n}\setminus A_{n})=0$. Let $B=\cup_{n}B_{n}$, which is a Borel set. Clearly $A\subseteq B$. Moreover, $B\setminus A=\cup_{n}(B_{n}\setminus A)\subseteq\cup_{n}(B_{n}\setminus A_{n})$, so $m(B\setminus A)=0$.

Danny Pak-Keung Chan
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  • Sorry, I overlook that you are working on $\mathbb{R}^n$ instead of $\mathbb{R}$. However, the technique is the same. – Danny Pak-Keung Chan Oct 29 '20 at 20:59
  • I think there is a typo in your definition of $A_n$. – Ruy Oct 30 '20 at 00:03
  • Fixed. Thank you. – Danny Pak-Keung Chan Oct 30 '20 at 00:45
  • Thanks, I have forgotten some stuff. For $m(A)<\infty$, why does there exist an open $U_n$ s.t. $A\subset U_n$ and $m(U_n)-m(A)<\frac{1}{n}$? For $m(A)=\infty$, we use $B_n$ to contain $A_n$ but then again may I know why we can find $B_n$ s.t. $m(B_n\setminus A_n)=0$? –  Oct 30 '20 at 01:29
  • Maybe I am missing something, but why can't we just use outer regularity to find a Borel set $B$ such that $A\subseteq B$ with $m(B\setminus A)=0$? Then the claim follows for indicator functions, hence for simple functions. – Matematleta Oct 30 '20 at 01:31
  • @Matematleta I think Danny is trying to prove outer regularity –  Oct 30 '20 at 01:32
  • @manifolded It's the definition of Lebesgue measure using the outer measure. If $m(A)=\infty$, intersect $A$ with a ball of radius $n$ and then extend by continuity of the measure. – Matematleta Oct 30 '20 at 01:33
  • @manifolded Yes, ok. – Matematleta Oct 30 '20 at 01:36
  • @Matematleta Thanks, I see. May I know how to prove this holds true for arbitrary $f$, I am guessing it does. I know that any Lebesgue measurable function can be approximated by a simple function. –  Oct 30 '20 at 01:37
  • @manifolded I will write up an answer shortly. – Matematleta Oct 30 '20 at 01:39
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    @Matematleta Thanks –  Oct 30 '20 at 01:40
  • @manifolded You may need to recall how we define Lebesgue measure. One approach is firstly defining Lebesgue outer measure $m^\ast$ on all subsets of $\mathbb{R}$. Then, Lebesgue measure is just the restriction of $m^\ast$ on the $\sigma$-algebra of Lebesgue measurable sets. – Danny Pak-Keung Chan Oct 30 '20 at 04:30
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Here is a proof for arbitrary Lebesgue measurable $f$. As the other answer shows, $A=F\cup N$ for some Borel set $F$ and a null set $N$. Let $U\in \tau_{\mathbb R}.$ Then the fact that $f$ is Lebesgue measurable implies that $f^{-1}(U)=V$ is Lebesgue measurable and so has the form $F\cup N$, as above.

Let $\{I_n\}_{n\in \mathbb N}$ be an enumeration of the collection of intervals with rational endpoints, which form a base for $\tau_{\mathbb R}.$ Then, there are Borel sets $\{F_n\}_{n\in \mathbb N}$ and null sets $\{N_n\}_{n\in \mathbb N}$ such that $f^{-1}(I_n)= F_n\cup N_n.$ Furthermore, $U=\bigcup^\infty_{k=1}I_{n_k}$ for some subsequence. Then, $f^{-1}(I_{n_k})=F_{n_k}\cup N_{n_k}.$

Define $g:\mathbb R^n\to \mathbb R$ by $g(x)=f(x)$ if $x\in F_n$ for some $n$ and $g(x)=0$ otherwise. Then, $g$ agrees with $f$ except perhaps on $\bigcup_n N_n$, which is null. And $g^{-1}(U)=g^{-1}(\bigcup^\infty_{k=1}I_{n_k})=\bigcup^\infty_{k=1}F_{n_k}\cup g^{-1}(\{0\}),$ which is a Borel set. It follows that $g$ is Borel measurable.

Remark: if you know that $\mathscr B(\mathbb R)$ is generated by the intervals $\{(-\infty, r):r\in \mathbb Q\}$, then you can show that it suffices to prove the claim for these intervals only. Then the proof is much easier: For each $r\in \mathbb Q,$ choose a Borel set $B_r\subseteq f^{-1}((-\infty,r))$ such that $f^{-1}((-\infty,r))=B_r\cup N_r$ for some null set $N_r$. Now define $g(x)=f(x)$ if $x\in B_r$ for some $r$ and $0$ otherwise. Then, $g^{-1}((-\infty,r))=B_r\cup g^{-1}(\{0\}).$

Matematleta
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  • Once we know that for each Lebesgue measurable set $A$, there is a Borel set $B$ such that $\mu(A\Delta B)=0$, then the remaining part is trivial:

    Let $f=\sum_{k=1}^{n}\alpha_{k}1_{A_{k}}$ be a simple Lebesgue measurable function, where $A_{k}$ is Lebesgue measurable. For each $k$, choose a Borel set $B_{k}$ such that $\mu(A_{k}\Delta B_{k})=0$. Define $g=\sum_{k=1}^{n}\alpha_{k}1_{B_{k}}$, then $g$ is a simple Borel function. Let $C=\cup_{k=1}^{n}(A_{k}\Delta B_{k})$. Then $m(C)=0$. It is routine to verify that $f(x)=g(x)$ for each $x\in C^{c}$.

     – Danny Pak-Keung Chan Oct 30 '20 at 03:23
  • May I know what is $\tau_{\mathbb{R}}$, is it the Borel sigma-algebra? –  Oct 30 '20 at 03:50
  • Also, how do we know that $g^{-1}({0})$ is Borel measurable? –  Oct 30 '20 at 04:00
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    $\tau_{\mathbb R}$ is the collection of open sets in $\mathbb R$ and $g^{-1}({0})=\left(\bigcup^\infty_{k=1}F_{n_k}\right)^c$ which is Borel – Matematleta Oct 30 '20 at 04:04
  • Matematleta's proof has a fundamental problem: The definition of $g$ relies on $U$ because $F_{n_k}$ depends on U. That is, for different choice of $U$, $g$ is different. Of course, this is not permitted. – Danny Pak-Keung Chan Oct 30 '20 at 04:26
  • This step should be easy as I showed on the above remark. – Danny Pak-Keung Chan Oct 30 '20 at 04:32
  • @DannyPak-KeungChan Pak I think it's ok, because when we choose $x$, if there are two open sets $U, V$ such that $x\in f^{-1}(U)$ and $x\in f^{-1}(V)$, it does NOT affect the definition of $g$ which only sends $x$ to $f(x)$ if there is SOME Borel set containing $x$, as in the definition There will be of course many of these or none. But in each case $x\mapsto f(x)$ if yes and to $0$ if not, so $g$ is well-defined. – Matematleta Oct 30 '20 at 04:35
  • @ Matematleta Consider two cases for choosing $U$. Case 1: $U=\emptyset$. In this case, $f^{-1}(U)=\emptyset$. Then all $F_{n}$and $N_{n}$ are forced to be the empty set. In this case, for each $x\in\mathbb{R}$, $x\notin F_{n}$ for any $n$. Hence $g(x)=0$.

    Case 2: $U=\mathbb{R}$. Then $f^{-1}(U)=\mathbb{R}$. In this case, unless $f=0$, there will be a lot of $x$ for which $x\in F_{n}$ for some $n$ and hence $g(x)=f(x)\neq0$.

    We can clearly see that the $g$ defined in Case 1 and Case 2 are different and depend on the choice of $U$ at the very beginning.

     – Danny Pak-Keung Chan Oct 30 '20 at 05:35
  • But the empty set is Borel being the complement of $\mathbb R$ so just add it to the collection of $F_n$. The fact that it is also null is irrelevant. Please see my reworked proof. I think it is correct. – Matematleta Oct 30 '20 at 05:41