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On a previous discussion found in Continuity of improper integral with a continuous integrand. Matematleta explains how to prove an improper integral is continuous given the fact that the integrand is continuous and that the integral is uniformly convergent. But I do not understand some details of his proof. I'll provide what I understand so far.

Let $$F(s) = int_{0}^{infty} f(s,t) dt$$

So I believe by the triangle equality and the linearity of the integral we have

$$left | F(s)-F(z) right | leq left | int_{0}^{M} f(s,t)-f(z,t) dt right | + left | int_{M}^{infty} f(s,t)-f(z,t) dt right |$$ $(1)$

Now since we know that $f(s,t)$ is continuous, that means that $left | f(s,t)-f(z,t) right | < epsilon$ whenever $d((s,t),(z,t)) < delta_{1}$, where $epsilon > 0$ and $delta_{1} > 0$. I have not really seen the notation of $d((s,t),(z,t))$ before, but from my research, I'm assuming it means a metric of distance, but is that right?

Continuing with the proof, we find that by the triangle equality we have

$$left | int_{0}^{M} f(s,t)-f(z,t) dt right |leq int_{0}^{M} left | f(s,t)-f(z,t) right | dt leq Mepsilon$$

I believe that this also comes from the triangle equality and the linearity of the integral, but we also have $$left | int_{M}^{infty} f(s,t)-f(z,t) dt right |leq left | int_{M}^{infty}f(s,t)dt right | + left | int_{M}^{infty} f(z,t) dt right |$$

Since we know that $F(s)$ is uniformly convergent, one can choose M such that $$left | int_{M}^{infty}f(s,t)dt right | < epsilon$$ $(2)$
$$left | int_{M}^{infty}f(z,t)dt right | < epsilon$$ $(3)$

Combining $(2)$ and $(3)$ together we have $$left | int_{M}^{infty} f(s,t)-f(z,t) dt right | < epsilon$$ $(4)$

Here is what I do not understand. If we were to combine $(1)$ and $(4)$, from my understanding we should have
$$left | F(s)-F(z) right | leq (M+1)epsilon$$
But isn't it supposed to be less than or equal to $epsilon$ to complete the proof? Also, I'm pretty sure that $delta$ is defined as $$left | s-z right |<delta$$ so according to the link I provided, why is $delta = delta_{1}/2$? The three questions I have listed are the things that I do not understand about this proof of how F(s) is continuous.

The point is that you can make your estimate of $left|int_0^infty (f(s,t) - f(z,t)); dtright|$ arbitrarily small. If you want it less than $epsilon$, change the $epsilon$ in the previous part of the proof to $epsilon' = epsilon/(M+1)$. Then you'll get $$ left|int_0^infty (f(s,t)-f(z,t)); dtright| < (M+1) epsilon' = epsilon$$
Similarly, you can use $delta_1$ instead of $delta$ if needed for the proof.

EDIT: BTW, from (2) and (3) you actually get
$$ left| int_M^infty (f(s,t) - f(z,t)); dt right| < 2 epsilon $$
not just $epsilon$. So use $M+2$ instead of $M+1$.

I don't see where Matematleta actually used $delta = delta_1/2$. You actually need the fact that continuous functions on a compact set are uniformly continuous, so that there is $delta > 0$ such that $|x - z| < delta$ implies
$|f(x,y) - f(z,y)| < epsilon'$ for all $y in [0,M]$.