# Show a function defined by an integral is Borel measurable.

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.

Hint: Consider the sequence of Borel measurable functions on $[0,1]$ given by $$f_n(x) = \sum_{k=1}^{n} f\left (x,\frac{k}{n}\right)\cdot \frac{1}{n}.$$
• 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? – S. D. ZHU Dec 15 '17 at 0:30