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Suppose $f : [a,b] \times [c,d]\to\Bbb{R}$ is continuous. Let $F:[c,d]\to \Bbb{R}$ and $$F(y)=\int_a^bf(x,y) \space\space dx.$$

Show $F(y)$ is continuous.

I try the following : Let $y,t \in [c,d]$. Suppose $F(y)$ is not continuous $\implies$ $F(y)$ is not uniformly continuous. We have that $\forall \delta>0\exists \epsilon >0$, s.t $$\vert y-t\vert<\delta\implies\vert F(y)-F(t)\vert=\Biggr\vert\int_a^bf(x,y)-\int_a^bf(x,t)\Biggr\vert=0\geq\epsilon.$$

A contradiction. I feel like this is wrong, since I didnt invoke the continuity of $f$ at any point. What should I take a look at to understand this problem?

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In your attempt, the statement:

$$ |F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|=0 $$ doesn't follow from what you assumed, that is, neither the form of $F$ nor that it is not uniformly continuous. Thus you can't get the contradiction $0\ge \epsilon$.


By definition of continuity of $F$, let $\epsilon >0$. We need to show there is some $\delta>0$ s.t for $y, t \in [c, d]$, $$ |y-t|<\delta \implies |F(y)-F(t)|<\epsilon $$

Since $f$ is continuous on $[a,b]\times[c,d]$, it is uniformly continuous there. Hence, in particular, given $\epsilon >0$, there is $\delta >0$ s.t. for all $x\in[a,b]$, $$ |(x,y) - (x,t)|=|y-t| <\delta \implies |f(x,y)-f(x,t)|<\epsilon/(b-a) $$

Thus: $$ |F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|\\ \leq \int_a^b \left|f(x,y)- f(x,t)\right| dx\\ \leq \epsilon(b-a)/(b-a) = \epsilon $$

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