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Statement: Let $f:[a,b]\times\mathbb{R} \to \mathbb{R}, (x,t) \mapsto f(x,t)\in\mathbb{R}$ be a continuous function. Then $$ F(t) := \int_a^b f(x,t)dx $$ is continuous.

I understand the proof for this up until the last part where continuity of $F$ is shown.

Let $t_0\in\mathbb{R}$ and $r>0$ and let $\epsilon>0$. Since $f$ is continuous, it is uniformly continuous in $X=[a,b] \times [t_0-r,t_0+r]$. Therefore we have that $\exists \delta = \delta(t_0,\epsilon)$ s.t.

$$ \|(x,t)-(y,s)\|_2<\delta \implies |f(x,t)-f(y,s)|<\epsilon $$ for $(x,t),(y,s) \in X$. Now let $h\in\mathbb{R}$ s.t. $|h|<\min\{r,\delta\}$. Then \begin{equation} |F(t_0+h) - F(t_0)| = \left|\int_a^b(f(x,t_0+h)-f(x,t_0))dx\right| \le (b-a)\epsilon \tag 1 \end{equation}

by using uniform continuity of $f$ in the integrand and noting that the arguments are less than $\delta$ apart.

Now, the part I am confused on is how this implies continuity of $F$ at $t_0$? I am slightly rusty on these topics.

I believe we can show $\lim_{h\to0}F(t_0+h)=F(t_0)$ and this is continuity of $F$ at $t_0$. How do I show this limit precisely from (1)? Intuitively my thoughts are that as $\epsilon$ is arbitrarily small, $h$ also gets smaller as $\epsilon$ is small.

Lucas
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1 Answers1

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First of all you have a typo: $F(t_0+h,t_0)$ should just be $F(t_0+h)$. And second, you’ve shown that for every $\epsilon>0$, there’s a $\delta’>0$ such that for all $|h|<\delta’$, we have $|F(t_0+h)-F(t_0)|\leq (b-a)\epsilon$.

This is one of the many equivalent ways of defining continuity of $F$ at $t_0$. If you’re worried about the final inequality being $\leq (b-a)\epsilon$ rather than $<\epsilon$, then that’s a simple exercise in analysis for you to figure out (there are loads of proofs online; see also this MSE answer of mine for about 12 different ways of saying something is a limit).

peek-a-boo
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  • Thanks. I understand clearly $(b-a)\epsilon$, not worried about that. My question is how to show this definition you've given is equivalent to the more standard definition. That definition being: $\forall \epsilon >0 \exists \delta>0 s.t. \forall t\in\mathbb{R}, |t-t_0|<\delta \implies |F(t)-F(t_0)|<\epsilon$. – Lucas May 27 '24 at 09:31
  • Also, is it rigorous enough to say that from $(1)$ that we have $\lim_{h\to0} |F(t_0+h)-F(t_0)| \le (b-a)\epsilon$?. Then since $\epsilon$ arbitrary we must have equality, that is, continuity of $F$. – Lucas May 27 '24 at 09:37
  • If $\epsilon$ is a positive number, then so is $\epsilon/2(b-a)$. Use this positive number in the definition of limits, to get a $\delta>0$ such that … Anyway this is a much more basic exercise than showing continuity of an integral; read the link and try to prove the various equivalences. Atleast try to prove one or two of them. If you really can’t then search closely on the site. – peek-a-boo May 27 '24 at 09:37
  • It’s not rigorous if you write $\lim\limits{h\to 0}$, because how do you know the limit even exists? If you want to argue in this manner, you would have to use $\limsup\limits_{h\to 0}$, and then after that take the limit $\epsilon\to 0^+$ (i.e squeeze theorem), and only then can you conclude the limit $\lim\limits_{h\to 0}$ exists and equals $0$. This approach with $\limsup$ is convenient but for here it’s overkill… you really should prove the 12 equivalences in the link before seriously moving on with analysis. – peek-a-boo May 27 '24 at 09:42
  • I see. So I understand clearly I have shown that $\forall \epsilon >0 \exists \delta>0$ s.t. $\forall h\in\mathbb{R}, |h|<\delta \implies |F(t_0+h)-F(t_0)| < \epsilon$. And so this is equivalent to $$ \forall \epsilon >0 \exists \delta>0 \text{s.t.} \forall t\in\mathbb{R}, |t-t_0|<\delta \implies |F(t)-F(t_0)| < \epsilon. $$ ?. – Lucas May 27 '24 at 09:45
  • Oh yes those are equivalent too (so now that gives us atleast 24 ways of describing limits). – peek-a-boo May 27 '24 at 09:45
  • Perfect, thanks. I understood this proof clearly and was just unneccesarily confused on this small part as I hadn't before seen this definition of continuity in a proof before. – Lucas May 27 '24 at 09:47
  • Well, it’s all part of an analyst’s toolbox, and now you know it :) – peek-a-boo May 27 '24 at 09:48
  • Also, I was already aware of those 12 equivalences you had linked, I understood this part clearly. As you can see my problem was simply realising that what I had shown was a reformulation of the equivalent 'standard' definition of continuity. This is separate to what you attached in the link. I don't think I would have made it this far without knowing those. – Lucas May 27 '24 at 09:50