Finding a sequence of integrable functions on $[0,1]$ which is pointwise convergent to zero whose integrals increase without bound

Question

As the title suggests I am looking for examples of sequence of integrable functions on $[0,1]$ which are pointwise convergent to zero. The sequence of integrals $\left\{\int_{0}^{1}f_n\right\}_{n=1}^{\infty}$ needs to increase without bound.

I've tried numerous integrable functions that are pointwise convergent to zero but I have yet to pinpoint one that satisfies the condition about integrals increasing without bound. Every single function I've found gives me smaller and smaller integrals. This is frustrating, to say the least.

Answer 1

$$f_n(x)= \cases{n^2 & if $\ \ 0<x <1/n$ \\ 0 & otherwise}$$

Answer 2

Take for $ n>0$, and $ x\in[0,1]$ :

$$f_n(x)=0 \;\;if\; \;x\ge \frac{1}{n^2},$$

$$f_n(x)=n^2e^n\; \; if\; \; 0<x<\frac{1}{n^2}$$

and $$f_n(0)=0$$

Check that $$\int_0^1f_n(x)dx=e^n$$

Answer 3

We can do this without piecewise defintions. Let $f_n(x)= n^3x^n(1-x).$ Then $f_n(x)\to 0$ at $x=0,x=1.$ On $(0,1),$ we have $f_n(x)\to 0$ pointwise, simply because exponential decay always beats out polynomial growth.

Does $\int_0^1 f_n$ increase to $\infty?$ Yes, because

$$\int_0^1 n^3x^n(1-x)\,dx = \int_0^1 n^3(x^n-x^{n+1})\,dx$$ $$ = n^3\left (\frac{1}{n+1}- \frac{1}{n+2}\right ),$$

which increases to $\infty.$