If $f_N $ is a function sequence that is integrable on $(a; b)$ for all $N $ and $\lim_{N\to\infty} f_N (x)$ exists for all $x$ in $(a; b) $
Does the following necessarily hold?: $$\lim_{N\to\infty} \int_{a}^{b} f_N (x) dx = \int_{a}^{b} \lim_{N\to\infty} f_N(x) dx$$
On
The statement is correct only if the functions $f_N(x)$ tend to $f(x)$ uniformly.
Proof: Let $f(x)=\lim_{n\to\infty}f_n(x)$. Uniform convergence implies that:$$\forall\epsilon>0\quad,\quad \exists N\quad,\quad\forall x\in(a,b), n>N\quad,\quad |f_n(x)-f(x)|<\epsilon$$which leads to $$f(x)-\epsilon<f_n(x)<f(x)+\epsilon$$since both $f_n(x)$ and $f(x)$ are integrable we can argue that:$$\int_{a}^{b}f(x)-\epsilon dx<\int_{a}^{b}f_n(x)dx<\int_{a}^{b}f(x)+\epsilon dx$$or$$\int_{a}^{b}f(x)dx-(b-a)\epsilon <\int_{a}^{b}f_n(x)dx<\int_{a}^{b}f(x) dx+(b-a)\epsilon$$which means that$$|\int_{a}^{b}f_n(x)dx-\int_{a}^{b}f(x) dx|<(b-a)\epsilon$$and implies $$\lim_{n\to\infty}\int_{a}^{b}f_n(x)dx=\int_{a}^{b}f(x) dx=\int_{a}^{b} \lim_{n\to\infty}f_n(x)dx$$
In general, no.
Take the function
$$f_N(x)=\begin{cases}N&\text{ if } 0<x<\frac1N\\ 0&\text{ if } x\geq \frac1N\text{ or } x=0\end{cases}$$
on $[0,1]$ as a counterexample.
Then, $$\int_0^1 f_N(x)dx = 1$$ for all $N$ and $$\lim_{N\to\infty} f_N(x) = 0$$ for all $x\in[0,1]$.