A question about sequences of integrable functions

Question

Say $f_n$ is some sequence of Lebesgue integrable maps $[0,1] \to \mathbb R$ such that for all $x\in [0,1]$:

$$ \lim_{n \to \infty} f_n (x) = 0$$

That is, the pointwise limit is the zero function.

Now consider the sequences

$$ \int_0^1 f_n(x)dx$$

and

$$ \int_0^1 |f_n(x)|dx$$

I believe that if the pointwise limit function is zero then these should also tend to zero but I'm not 100% sure.

Especially, I'm not sure how to argue it because they are only integrable and not necessarily continuous.

Does anyone have any insights into this?

Answer 1

Nope, not true, even for everywhere pointwise convergence: Consider $$ f_n(x) = \begin{cases} n^2x & 0\leq x\leq 1/n\\ 2n-n^2x & 1/n<x\leq 2/n\\ 0 & x>2/n \end{cases}. $$ (Note that all $f_n$ are continuous…). Also look up Lebesgue's dominated convergence theorem.

Answer 2

The answer is no. Here is a counterexample that $f_n(x)\to0$, but $\int_0^1 f_n(x)dx\to\infty$ as $n\to\infty$.

Let $f_n(x)=n^2xe^{-nx^2}$ on $[0,1]$. Then for any $x\in[0,1]$ $$ \lim_{n \to \infty} f_n (x) = 0 $$ But $$ \int_0^1 f_n(x)dx=\frac{n}{2}(1-e^{-n})\to\infty $$

Answer 3

An easy counterexample in the case of almost everywhere convergence, since I've already written it. Dirk took care of the question better.

I also recommend to read something like this Tao's post about modes of convergence, to get a better idea of what's going on.

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Take $f_n(x)=n\chi_{[0,\frac{1}{n}]}(x)$. Then $f_n \to f$ almost everywhere and $$ \int_0^1 |f_n(x)|\,dx=\int_0^1 f_n(x)\,dx=n\int_0^1 \chi_{[0,\frac{1}{n}]}(x)\, dx= n \mu\left(\left[0,\frac{1}{n}\right]\right)=1, $$ for any $n$ and where we denoted by $\mu$ the Lebesgue measure.

Answer 4

Or look at $f_n(x) = n^2x^n(1-x).$