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Prove that if $(f_n)$ is a sequence of Borel measurable functions and if $f(x)=\lim_{n\to \infty}f_n(x)$ exists in $\mathbb{R}$, then $f$ is Borel measurable. In fact $f$ is Borel measurable even if we only have $f(x)=\lim_{n\to\infty}$ almost everywhere on $D$, some measurable domain.

Attempt/Thoughts:

Suppose $(f_n)$ is Borel measurable. That is, the preimage of $f_n$ for every interval of the form, (for $\alpha\in\overline{\mathbb{R}})$, $((\alpha,\infty])$ is a Borel set. We are given pointwise convergence, so by definition, $\forall\epsilon>0\exists N\in\mathbb{N}: \forall n\geq N: |f_n(x)-f(x)|<\epsilon$

I'm not sure how to proceed from here.

I'm not sure how to start the second part at all. I know that it means that $f$ is Borel measurable even if $f(x)=\lim_{n\to\infty}f_n(x)$ on $D\setminus E$, where $m(E)=0$.

Any suggestions would be welcome. Thanks.

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I strongly suspect that the second part is not true. This because $(\mathbb R,\mathcal B,\lambda)$ where $\mathcal B$ denotes the Borel sigma algebra and $\lambda$ the Lebesgue measure, is not a complete measure space.

That makes it possible to find sets $D\subset E\subset\mathbb{R}$ such that $D$ is a Borel set, $E$ is not a Borel set and $\lambda\left(\mathbb{R}-D\right)=0$.

Define $f=1_{E}$ and for $n=1,2,\dots$ define $f_{n}=1_{D}$.

Then $\lim_{n\rightarrow\infty}f_{n}\left(x\right)=f\left(x\right)$ on $D$, hence almost everywhere. However $f$ is not Borel-measurable.

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  • $\begingroup$ I think you meant to write "That makes it possible to find sets $E \subset D \subset \mathbb R$ such that $D$ is a Borel set, $E$ is not a Borel set [...]". For example with $D$ being the Cantor set which has $2^{\mathfrak c}$ subsets, but only $\mathfrak c$ many subsets of $\mathbb R$ are Borel (although the proof of this fact is usually omitted in a course on measure theory). $\endgroup$ – kahen Mar 2 '15 at 10:44
  • $\begingroup$ @kahen No. What I meant is: take some Borelset $B$ with measure zero that has a non Borel subset $C$ (that is possible). Then let $D$ be the complement of $B$ and let $E$ be the complement of $C$. The fact that $D\subset E$ garantees that $f_n(x)\rightarrow f(x)=1_E(x)$ on $D$, wich is essential here. My set $E$ is not referring to the set $E$ in the question. $\endgroup$ – drhab Mar 2 '15 at 10:52
  • $\begingroup$ @drhab: If you don't mind, could you explain a little more about how you know that the the choice of $f_n$ above results in $f$ not being Borel measurable? And just to clarify, $f_n=1_D$ doesn't have an index so the sequence of functions is just the same function, $1_D$, for the entire sequence? $\endgroup$ – Sujaan Kunalan Mar 13 '15 at 23:52
  • $\begingroup$ $f_{n}=1_{D}$ for each $n$ guarantees that for every $x\in D$ and every set $E$ with $D\subseteq E$ we have $f_{n}\left(x\right)=1_{D}\left(x\right)=1=1_{E}\left(x\right)$. This is even stronger than $\lim_{n\rightarrow\infty}f_{n}\left(x\right)=1_{E}\left(x\right)$. Then $\lambda\left(\mathbb{R}-D\right)=0$ tells us that $f_{n}$ converges to $1_{E}$ almost everywhere. Do you agree with that? Secondly set $E$ can be chosen such that $1_{E}$ is not a Borel-measurable function. This because the measure space $\left(\mathbb{R},\mathcal{B},\lambda\right)$ is not complete. Do you agree with that? $\endgroup$ – drhab Mar 14 '15 at 9:57
  • $\begingroup$ Actually you cannot say that "the choice of $f_n$ result in $f$ not being Borel measurable". This because the choice of $f_n$ is not completely determining for $f$. There are more functions $f$ that satisfy $f_n\rightarrow f$ a.e.. The point is that some of these functions are not Borel-measurable, so that from $f_n\rightarrow f$ a.e. you cannot conclude that $f$ is Borel-measurable. $\endgroup$ – drhab Mar 14 '15 at 10:05
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Part (1):

Equivalently, $f_n$ is Borel measurable if $\{x : f_n(x) < \alpha\} \in\mathcal{B}$ for every $\alpha$.

If $f(x) < \alpha$ (for some $x$), then there exists $m \in \mathbb{N}$ such that $f(x) < \alpha - 1/m$. Since $f_n(x) \to f(x)$ it follows that there exists $N \in \mathbb{N}$ such that $f_n(x) < \alpha - 1/m$ for $n \geqslant N$. The converse is true as well.

Since $f_n$ is Borel measurable $\{x:f_n(x) < \alpha - 1/m\} \in \mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra we have

$$\{x: f(x) < \alpha\} = \bigcup_{m=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n= N}^\infty\{x:f_n(x) < \alpha - 1/m\} \in \mathcal{B},$$

and $f$ is Borel measurable.

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