Q:

Prove that $F$ is continous

Let $f:[0,1] \times \mathbb{R} \to \mathbb{R}$ continous function and consider $F(y)= \int_0^1 f(x,y) dm(x)$. I want to show that $F$ is continous. 
I tried: Take $y' \in \mathbb{R}$. Let $(y_n)_n$ any sequence such that $y_n \to y'$. I have to show that $\lim F(y_n)=F(y')$. For that i want to use some convergence theorem. Define $f_n(x)=f(x,y_n)$ using continuity of $f$ then:
$$ \lim_{n \to \infty} f_n(x)= \lim_{n \to \infty} f(x,y_n)= f(x,y)$$
I can not finish. I can not limit the sequence $f_n$.

A:

Fix $y \in \mathbb{R}$. Note that $f$ is uniformly continuous in the compact set $K:= [0,1] \times [y-1,y+1]$. Now, let $\varepsilon > 0$.
By uniform continuity in $K$, there exists $\delta > 0$ such that if $\|(x,y)-(x',y')\| < \delta$ and $(x,y),(x',y') \in K$ then $|f(x,y)-f(x',y')| < \varepsilon/2$.  Without loss of generality, we can additionally assume $\delta < 1$. 
If $|y-y'| < \delta$, then we get that $\|(x,y)-(x,y')\| < \delta$ for each $x \in [0,1]$. Moreover it has to be $(x,y),(x,y') \in K$ because $|y-y'| < \delta < 1$. This says that
$$
|f(x,y)-f(x,y')| < \varepsilon/2 \quad (\forall y' \in B_\delta(y), x \in [0,1])
$$
and thus
$$
|F(y)-F(y')| \leq \int_0^1|f(x,y)-f(x,y')|d\mu \leq \int_0^1\varepsilon/2 \cdot d\mu = \varepsilon/2 < \varepsilon.
$$
which concludes the proof: for a given $\varepsilon$, we have found $\delta > 0$ such that if $|y-y'| < \delta$, then $|F(y)-F(y')| < \varepsilon$.

