Condition for $\int_{a}^{b}f_n(x)dx\rightarrow\int_{a}^{b}f(x)dx$

Question

I know that the condition needed for $\int_{a}^{b}f_n(x)dx\rightarrow\int_{a}^{b}f(x)dx$ is for $f_n$ to converge uniformly to $f$. However, I'm unable to come up with a sequence of functions $f_n:[a,b]\rightarrow\mathbb R$ that converges pointwise to $f:[a,b]\rightarrow \mathbb R$ but not uniformly such that the above does not hold. Does anybody know any relatively simple functions that show pointwise convergence is not sufficient? (Preferably not piecewise).

EDIT: where $f_n,f$ are both Riemann integrable (for all $n$)

Answer 1

On $[0,1]$ let $f_n(x) = n^2x^n(1-x).$ Then $f_n(x)\to 0$ pointwise on $[0,1].$ But a straightforward computation shows $\int_0^1 f_n(x)\, dx \to 1.$

Answer 2

Let $a=0$, $b=1$. Pick your favorite enumeration $(r_i)$ of the rarional numbers on $[0,1]$. Define the sequence as $f_0=0$ and $f_n(r_i)=1$ if $i\leqslant n$ and $0$ otherwise (for example, $f_5$ is $1$ at the first $5$ rational numbers, and zero otherwise). It concerges pointwise to the indicator function of the rarionals on $[0,1]$, but it's not Riemann integrable.

Edit: Arzela's dominated convergence theorem for Riemann integrals (https://sites.math.washington.edu/~morrow/335_15/dominated.pdf) states that if $\exists M>0$ so that $|f_n|<M$, $f_n \to f$ pointwise and all of the functions are Riemann-integrable then $$\lim_n \int f_n = \int f $$ As you can see in the other answers, you can construct a counterexample if the sequence is not uniformly bounded.

Answer 3

Consider the functions on the interval $[0,1]$ (you can generalize easily to any interval) as follows$$f_n(x)=\begin{cases}n&0<x<\frac1n&\\0&x\geq\frac1n\text{ or }x=0\end{cases}$$$$f(x)=0$$

Clearly, $\lim_{n\to\infty}f_n=f$, but $\int_0^1 f_n=1$ for all $n$ and $\int_0^1 f=0$.

Answer 4

Take $$f_n(0)=0$$ $$f_n(x)=3n^2 \text{ if } 0< x\le \frac{1}{n^2}$$

$$f_n(x)=0 \text{ if } x> \frac{1}{n^2}$$

Clearly, the sequence $(f_n)$ converge pointwisely to zero function at $[0,1]$.

the convergence is not uniform since $$\sup_{[0,1]}|f_n-0|=3n^2\to +\infty$$ but $$\forall n>0\;\; \int_0^1f_n=3 $$ and $$\lim_{n\to+\infty}\int_0^1f_n=3 \ne \int_0^10dx$$