Is the Pointwise Limit of Measurable Functions Also Measurable?

[jose80]

Hi,

I got across this question, if $F:[0,1] \to \mathbb{R}$ is differentiable, then how to show it is derivative $F'$ is Borel measurable?

Any idea?

 

[Landau]

Hi jose80, you will need to use [tex ]..[ /tex] (without the spaces) instead of $..$.

Write F' as the (pointwise) limit of measurable functions, using the definition of derivative.

 

[g_edgar]

In more detail, for each fixed $n$ the function $f_n$ defined by
$$f_n(x) = \frac{f(x+1/n)-f(x)}{1/n}$$
is continuous, and $f'(x)$ is the pointwise limit.

 

[Landau]

@g_edgar: I purposely avoided this amount of detail, because it sounds like homework, and my hint seemed quite reasonable to me.

 

[jose80]

Hi, thanks for the answers, I tried to look up why a pointwise limit of continuous functions is Borel measurable, but I couldn't figure out that?

Any reference or hint?

 

[Landau]

The pointwise limit of (real-valued) measurable functions is measurable. That is one of the most basic and important results in (elementary) measure theory. If that's not in the book you're reading, then I'm pretty sure that's not a book about measure theory :)

E.g. see here.