I was reading the following proof:
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:
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.