One definiton of Borel measurable functions is:

The collection of borel measurable functions is the smallest collection of real valued functions on $\mathbb{R}$ that contains the continuous functions and is closed under pointwise limits.

Now my question. Is every Borel measureable function then according to this definition the pointwise limit of continuous functions?

What can go wrong is this: $\{f_n^k\}_{n=0}^{\infty}\rightarrow f^k$, and we assume that all $f_n^k$ are continuous. $f^k$ need not be continuous. And lets say that $\{f^k\}\rightarrow f$, now since the borel measurable functions is closed under pointwise limits, f has to be borel-measurable. But we haven't shown that it is a limit of continuous functions. Can such an f exist that is not a limit of continuous functions?

I tried proving that indeed f was a limit of continuous functions, but I couldn't finish my proof, I ran into this problem: Is this statement about limits true. I tried proving it with a diagonal-argument, but maybe there is another way to prove what we want?

Is the answer yes and if so, how do I prove it? Or is it not true, and is there a counter-example?


Tha answer is no.

Every pointwise limit of continuous functions is of Baire class $1$ (this is just the definition), and this implies that the set of discontinuities of $f$ is not all of $\Bbb{R}$ by a Baire category argument (see this post Construction of a function which is not the pointwise limit of a sequence of continuous functions).

But for example the Borel function $\chi_{\Bbb{Q}}$ (indicator function of the rationals) is discontinuous everywhere, hence not a pointwise limit of continuous functions.


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.