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

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 $$n$$)

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

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

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 $$\int_0^1 f_n(x)\, dx \to 1.$$

• I'm pretty sure you've see something like this before. Very much like showing $n^2/e^n\to 0.$
– zhw.
Commented Nov 14, 2019 at 19:44

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 $$\lim_{n\to+\infty}\int_0^1f_n=3 \ne \int_0^10dx$$

• This is literally just my answer. Commented Nov 14, 2019 at 19:06
• @DonThousand Okay, i changed some lines to get an other example. Commented Nov 14, 2019 at 19:12

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 $$|f_n|, $$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.