5
$\begingroup$

Let $f: \left [ 0, 1 \right ] \times \left [ 0, 1 \right ] \rightarrow \mathbb{R}$ satisfy:

  1. $f_x\left ( y \right ):= f\left ( x, y \right ) : \left [ 0, 1 \right ]\rightarrow \mathbb{R} $ is Riemann integrable;
  2. $f_y\left ( x \right ):= f\left ( x, y \right ) : \left [ 0, 1 \right ]\rightarrow \mathbb{R} $ is Borel measurable.

Then prove that $g(x):=\int_{\left [ 0, 1 \right ]} f_x\left ( y \right ) dy$ is Borel measurable.

I tried at the first consider Fubini theorem. However, it doesn't work due to the condition failure. Later I tried to prove directly from the definition of Borel measurability, and it does not go anywhere further.

Also, $f$ itself may not be measurable in the product measure; and Riemann integrability does not imply Borel measurability either.

Would you please give me some hints or ideas? Thank you.

$\endgroup$
1
$\begingroup$

Hint: Consider the sequence of Borel measurable functions on $[0,1]$ given by $$f_n(x) = \sum_{k=1}^{n} f\left (x,\frac{k}{n}\right)\cdot \frac{1}{n}.$$

$\endgroup$
  • $\begingroup$ I believe you are suggesting the pointwise limit of $f_n (x)$ is well defined since it is Riemann integrable and the limit is measurable as each term is. Am I right? $\endgroup$ – S. D. ZHU Dec 15 '17 at 0:30
  • $\begingroup$ Right, where "measurable" means "Borel measurable". $\endgroup$ – zhw. Dec 15 '17 at 3:00

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.