I'm having trouble with the question: Let $f: X \times [a,b] \to \mathbb{R}$ be a continuous application, with $X \subset \mathbb{R}^N$, and $\phi : X \to \mathbb{R}$ defined by: $\phi(x) = \int_{a}^{b}f(x,t)dt$ Show that $\phi$ is continuous.
I want to show that, given $\epsilon > 0$, $\exists \delta_{x,\epsilon} > 0$ so that $|y - x| < \delta \Rightarrow |\phi(y) - \phi(x)| < \epsilon$
So I chose $\frac{\epsilon}{b-a} > 0$, then $\exists \delta_{x,\epsilon} > 0$ so that $|y - x| < \delta \Rightarrow |f(y,t) - f(x,t)| < \frac{\epsilon}{b-a}$ as $f$ is continuous.
Then I have $f(x,t) - \frac{\epsilon}{b-a} < f(y,t) < f(x,t) + \frac{\epsilon}{b-a}$
Now I don't know how to proceed. I was really tempted to integrate everything, but inequalities would hold only if $f$ was uniform continuous, which is true within $t \in [a,b]$, as it's continuous in a compact, but not necessarily true for $x \in X$.
Could anyone help me with that? thank you.
Hint: continuity is a local property. It suffices to fix $x_0\in X$ and show $\phi$ is continuous at $x_0$. If you let $K$ be a compact neighborhood of $x_0$ all of your problems disappear.