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I was reading the following proof: enter image description here What I don't understand, is why we need $(f_n)$ to converge uniformly to $f$. In the proof they use the continuity of $f$ to justify the integrability. But why can't we rephrase the theorem as follows:

Let $(f_n)$ be a sequence of integrable functions on $[a,b]$, and suppose that $f_n\to f$ pointwise, with the limit function $f$ being integrable on $[a,b]$. Then...

If this is true, then I would say that theorem 25.2 is basically a corollary of the more general theorem I proposed.

Could someone tell me if my general paraphrase of the theorem is correct? Or do we really need uniform convergence?

If needed, this is the bit they refer to in the proof:

enter image description here

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    Did you take a look at Discussion $25.1(a)$? – 2017-02-05
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    @OpenBall Why is it relevant? 25.1(a) doesn't have anything to do with uniform convergence. – 2017-02-05
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    Can you prove that $g:=|f_n - f|$ is $\le$ that $h$ without uniform convergence? – 2017-02-05
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    @OpenBall Ah, I see. So otherwise we would need to pick an $N$ for each $x\in[a,b]$. Oooh, and I understand another thing: when integrating, we basically pick all $x\in[a,b]$ as a matter of speech; we need some $N\in\mathbb N$, such that $|f_n-f|$x\in[a,b]$, so that we can integrate 'properly'; i.e., having this necessary inequality. If my wordings aren't clear, don't mind, I understand it now. Thank you! – 2017-02-05

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The uniform convergence $f_n \to f$ garants that you can estimate $|f_n(x)-f(x)|$ uniformly by $\frac{\varepsilon}{b-a}$ that is for all $x\in [a,b]$.

You really need the uniform convergence. Have a look at $f_n\colon [0,2)\to \mathbf R$ with $$f_n(x)=\begin{cases}n^2x&0\leq x\leq 1/n\\2n-n^2x&1/n\leq x\leq2/n\\0&x\geq2/n.\end{cases}$$ Then $f_n \to 0$ pointwise (and not uniform) but $$1=\lim_{n\to\infty}\int_0^2f_n(x)\,\mathrm dx\ne\int_0^2\lim_{n\to\infty}f_n(x)\,\mathrm dx=0.$$ This easy example shows that you can not soften the premise of uniform convergence.

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    A slight remark (perhaps technical): uniform convergence can in fact be softened; see e.g. Lebesgue's monotone and dominated convergence theorems. – 2017-02-06
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    @Open Ball Thank you, you are right. I didnt mean to mention it because I think that the questioner is not familar with measure theory. – 2017-02-06