$\sqrt{n} \mathbb{1}_{[0,\frac{1}{n}]}$ converges simply almost everywhere to $0$.

Question

I’m currently reading a book about Fourier series and I stumbled about a statement that won’t leave me alone (It’s not in relation with a Fourier series though). The statement is as follows:

$\sqrt{n} \mathbb{1}_{[0,\frac{1}{n}]}$ converges simply almost everywhere to $0$.

I know that the indicator function on $[0,\frac{1}{n}]$ converges to $0$ as $n$ goes to $\infty$ but why does the product converges to 0? How do you show it properly?

Answer 1

If $x > 0$ then there exists $N \in \mathbf N$ with $\frac 1N < x$ and thus the property that $\sqrt{n} 1_{[0,\frac 1n]}(x) = 0$ for all $n \ge N$.

If $x = 0$ the sequence diverges to $\infty$.

If $x < 0$ the sequence is identically $0$.

Answer 2

Pick any real number $x \ne 0$. For sufficiently large $n$, the number $x$ will lie outside of the interval $[0, 1/n]$, so $\sqrt{n} \mathbb{1}_{[0,1/n]}(x) = \sqrt{n} \cdot 0 = 0$ for all large $n$.