What's the difference between $\int_{a}^{b} \lim_n f_{n}$ and $\lim_{n \to \infty} \int_{a}^{b} f_{n}$?

Question

I just started studying power series and I found this theorem.

Let be $\{f_{n}\}_{n}$ a sequence of integrable functions on $[ a,b]$ that uniformly converges to $f$ on $[a,b]$. If $f$ integrable on $[a,b]$ then $$\int_{a}^{b} f = \lim_{n \rightarrow \infty } \int_{a}^{b} f_{n}$$

And using the hypothesis, we should have

$$\int_{a}^{b} \lim_{n \rightarrow \infty } f_{n} = \lim_{n \rightarrow \infty } \int_{a}^{b} f_{n}$$

but I don't know what is the difference between write $\int_{a}^{b} \lim_n f_{n}$ or $\lim_n \int_{a}^{b} f_{n}$, for me those expression are the same.

Answer 1

If we don't have uniform convergence, those expressions may be different.

Consider this weird sequence of functions defined on $(0,1)$:

$$f_n(x)=\begin{cases}n &\text{for } x \in (0,\frac1n)\\0 &\text{otherwise}\end{cases}$$

For each $x \in (0,1)$, eventually $x > \frac1N$ for some large $N$.

Therefore $f_n(x)$ converges pointwisely to $f(x)\equiv 0$. This gives $\int_0^1 \lim_n f_n(x)dx = \int_0^1f(x)dx = 0$.

However, for each $n$, we have $\int_0^1 f_n(x)dx = \int_0^{1/n} n dx = 1$, so $\lim_n \int_0^1f_n(x)dx = \lim_n 1=1\ne0$.

Answer 2

Another example, let $f_n(x)=nx^n$ in $[0,1]$. Then $$ \lim_{n\to\infty}f_n(x)=\left\{\begin{array}{ll} 0&\text{ if }x\in[0,1),\\ \infty&\text{ if }x=1, \end{array}\right.\int_0^1f_n(x)dx=\frac{n}{n+1} $$ But $$ \int_0^1\lim_{n\to\infty}f_n(x)dx=0\neq\lim_{n\to\infty}\int_0^1f_n(x)dx=1. $$