Can you help me formally prove that $f_n:\mathbb R \to \mathbb R$,$f_n=n\chi_{[\frac{1}{n},\frac{2}{n}]}$ $\forall n\in\mathbb N$ converges pointwise to $f\equiv 0$ and that every $f_n$ is Lebesgue integrable?
It is not a homework,I am solving exercises for my tomorrow's measure theory exam.I thank you all in advance !
Fix any $x\in\mathbb{R}$. If $x> 0$, then there exists $N$ such that for all $n\geq N$ $x\notin[\frac{1}{n}, \frac{2}{n}]$ (because $\frac{2}{n} < x$) and thus $f_n(x)=0$. . And if $x \leq 0$, $f_n(x)=0$ for all $n$ anyway, as the support of $f_n$ is in $(0,\infty)$.
So for all $x\in\mathbb{R}$, $f_n(x) \xrightarrow[n\to\infty]{}0$; hence $(f_n)_n$ converges pointwise to $\bar{0}$ on $\mathbb{R}$.
Further, for any fixed $n$, $f_n$ is a simple function, and is thus Lebesgue integrable because $\mathbf{1}_{[\frac{1}{n}, \frac{2}{n}]}$ is measurable and has finite measure.