Dominated convergence, and continuity of an integral function Let $f:[0,1]\times [0,1]\to [0,1]$ be a measurable function with the properties:
$\color{red}{(1)}$ for every $x\in [0,1]$ is $y\mapsto f(x,y)$ continuous
$\color{blue}{(2)}$ for every $y\in [0,1]$ is $x\mapsto f(x,y)$ continuous
Prove the continuity of $g:[0,1]\to\mathbb{R}, g(x)=\int_0^1 f(x,y)\, dy$ on $[0,1]$.
I am currently working on this question.
To prove continuity of $g$ I want to show that for every given $x$ and every sequence $(x_n)$ converging to $x$, we have $\lim_{n\to\infty} g(x_n)=g(x)$.
So a proof should go along the line of:
$\lim_{n\to\infty} g(x_n)=\lim_{n\to\infty} \int_0^1 f(x_n,y)\, dy\stackrel{\text{DCT}}{=}\int_0^1\lim_{n\to\infty} f(x_n,y)\, dy\stackrel{\color{red}{(1)}}{=}\int_{0}^1 f(x,y)\, dy = g(x)$.
DCT referes to the dominated convergence theorem.
So this seems to be the only thing I have to verify.
It is easy to see that $f$ is integrable, as it is bounded by $1$. So $1$ works as an integrable upper bound.
We need a sequence of integrable functions, converging pointwise to $f$.
So lets fix $x$ and take a sequence $(x_n)$ converging to $x$.
Let $g_n(y)=f(x_n,y)$. Then $g_n(y)$ is a sequence of integrable functions, as said above, and
$\lim_{n\to\infty} g_n(y)=\lim_{n\to\infty} f(x_n,y)=f(x,y)$ by property $\color{blue}{(2)}$.
So we can apply the dominated convergence theorem as desired.
Do you mind checking this proof?
Thanks in advance for your comments.
 A: Let $I = [0,1]$. To show that $g: I\to I$, $x\mapsto \int_0^1 f(x,y)\, dy$ is continuous, you only need continuity in the first variable, for every $y\in I$. That is, you only need $\color{blue}{(2)}$ and not $\color{blue}{(1)}$.
For the following step that you have written in your post, you need to replace $\color{blue}{(1)}$ by $\color{blue}{(2)}$.

$\lim_{n\to\infty} \int_0^1 f(x_n,y)\, dy\stackrel{\text{DCT}}{=}\int_0^1\lim_{n\to\infty} f(x_n,y)\, dy\stackrel{\color{red}{(1)}}{=}\int_{0}^1 f(x,y)\, dy = g(x)$.

To better understand this, define for every $y\in [0,1]$, a function $f_y: I\to I$ given by $f_y(x):= f(x,y)$. The condition $\color{blue}{(2)}$ can be rephrased as the continuity of the family of functions $\{f_y\}_{0\le y\le 1}$. In other words, if $x_n\to x$, then $f_y(x_n)\to f_y(x)$. Coming back to your notation, this is just $\lim_{n\to\infty} f(x_n,y) = f(x,y)$.

Remark.
I am sure you know this, but to use DCT, you would define functions $\{f_n\}_{n=1}^\infty : I \to I$ by $f_n(y) = f(x_n,y)$ for a given sequence $\{x_n\}_{n=1}^\infty \subset [0,1]$ converging to $x$. Clearly, $|f_n(y)| \le 1$ for all $n$, and all $y\in I$, so DCT is applicable.

Remark $2$. You can use $\color{blue}{(1)}$ to show the continuity of $h:I\to I$ defined by $h(y):= \int_0^1 f(x, y)\, dx$. Take this as an exercise. Also, try to generalize this result to $n$-dimensions, i.e. functions $f:I^n\to I$ satisfying some similar hypothesis.
