How do I use Lebesgue's theorem to find $\lim_{n \to \infty} \int_{\{|f(t)|\le1/n\}}f(t)^2 dt$? As a final step of a functional analysis exercise, we arrive to the following limit integral that is shown to go to zero using Lebesgue's dominated convergence theorem:
$$\lim_{n \to \infty} \int_{\{|f(t)|\le1/n\}}f(t)^2 dt=\lim_{n \to \infty} \int_0^1nf(t)^2 1_{\{|f(t)|\le1/n\}}dt$$
To use the theorem we need a bounding function independent from n to bound the integrand:
$$nf(t)^2 1_{\{|f(t)|\le1/n\}} \le n(1/n^2) \le 1$$
and the pointwise limit of the integrand:
$$nf(t)^2 1_{\{|f(t)|\le1/n\}} \to 0$$
I don't understand how to bound the function and how to find that limit.
It looks like I may get an indeterminate form in the limit:
Since when $n\to \infty$, $ |f|\le 0 $inside the indicator set:
$\lim_{n \to \infty} nf(t)^2 1_{\{|f(t)|\le1/n\}} = \infty 0^21$
Anyway this type of calculation looks pretty informal
Can someone please explained it  in a-step-by-step rigourous way?
 A: You're close! The main thing to do is wait until the last second to actually evaluate your limit. Try to simplify the stuff you're limiting first, as otherwise you get meaningless quantities like "$\infty 0^2 1$" (which at least you recognize as meaningless).
First, the bound. I'm not sure where your misunderstanding is, since what you've written appears fine to me. To apply the dominated convergence theorem, we need to know that there is a dominating function $g$ so that

*

*$|f_n| \leq g$ for every $n$

*$\int |g| < \infty$
We're interested in the sequence $f_n = n f^2 1_{\{|f(t)| \leq \frac{1}{n}\}}$. We want to upper bound this by some function $g$. Well, let's follow our nose:
$1_{\{\text{blah}\}} \leq 1$ always, so it will never make $f_n$ bigger. But it's extremely useful because it lets us restrict attention to the set where $|f(t)| \leq \frac{1}{n}$ (do you see why?). So we get
$$
|f_n| = n |f|^2 1_{\{|f(t)| \leq \frac{1}{n}\}} \leq n \left (\frac{1}{n} \right )^2 1 = \frac{1}{n}
$$
So we always have $|f_n| \leq \frac{1}{n}$ on $[0,1]$. We want a bound $g$ that doesn't depend on $n$, so we take $g = \frac{1}{1} = 1$, since that bounds everything.
Now we have condition $1$ from before. $|f_n| \leq g$ for each $n$. We also have condition $2$, since we're working on $[0,1]$ and $\int_0^1 g \ dx = \int_0^1 1 \ dx = 1 < \infty$.
Then we can apply the dominated convergence theorem, and we see
$$\lim_{n \to \infty} \int_0^1 f_n \ dx = \int_0^1 \lim_{n \to \infty} f_n \ dx$$
Of course, what is $\lim_{n \to \infty} f_n$? Well we showed earlier that $|f_n| \leq \frac{1}{n}$ (while we were looking for an upper bound). Taking $n \to \infty$ shows
$$\lim_{n \to \infty} |f_n| \leq \lim_{n \to \infty} \frac{1}{n} = 0$$
So we must have $\lim_{n \to \infty} |f_n| = 0$ (by the squeeze theorem, if you like, since $0 \leq |f_n|$ for each $n$). Now by continuity of $|\cdot|$, this means $\left | \lim_{n \to \infty} f_n \right | = 0$
But the only way to have $|\text{blah}| = 0$ is if $\text{blah}$ itself is $0$. So we see that $\lim_{n \to \infty} f_n = 0$, which is exactly what you were trying to show.
Usually we don't go into anywhere near this much detail, but it's important to sometimes see all the moving parts laid out in front of you.

I hope this helps ^_^
A: Consider the cases $|f(t)| \leq \frac  1n$  and$|f(t)| >\frac 1n$. In  the first case $nf^{2}(t)1_{|f(t)| \leq \frac 1 n}=nf^{2}(t) \leq n(\frac  1{n^{2}})=\frac  1n$.  What happens in the second case?
A: Let $h_n(t)=nf(t)^2 1_{\{|f|\leq \frac 1 n\}}(t)$.
For the Lebesgue Dominated Converge theorem to apply, you need

*

*An integrable function $g\geq 0$ that dominates $h_n$ for all $n$. That is $|h_n|\leq g$ almost everywhere.

*The limit pointwise to exist: $h_n\rightarrow h$ almost everywhere for some integrable function $h$.

The key observation is that

*

*if $t$ is such $|f(t)|\leq \frac 1 n$, then $|h_n(t)|\leq n\frac 1 {n^2} = \frac 1 n$.

*if if $t$ is such $|f(t)|> \frac 1 n$, then $1_{\{|f|\leq \frac 1 n\}}(t)=0$ and thus $h_n(t)=0$.

So the value of $h_n(t)$ is either bounded by $\frac 1 n$, or is equal to $0$.  Therefore you can pick $g(t)=1$ since it will work in both cases. Also, the same reasoning implies that pointwise, $h_n(t)$ tends to $0$.
