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.


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.


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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.