If we define $\chi_{[0,\frac{1}{k}]}$ to be a characteristic function such that
$\chi_{[0,\frac{1}{k}]}(x)=\begin{cases} 1 ~~~ \text{if $x\in [0,\frac{1}{k}]$} \\ 0 ~~~ \text{otherwise.} \end{cases}$
How do we show that the function $k\chi_{[0,\frac{1}{k}]}$ does not converge to a function $f$ point wise so that $\int |f_k-f| \rightarrow 0$ as $k\rightarrow \infty$? I am lost since when $k\rightarrow \infty$ we have $\chi_{[0,\frac{1}{k}]}\rightarrow \chi_{[0,0]}$. The part that confuses me the most is if its true that $k\chi_{[0,\frac{1}{k}]}\rightarrow \infty \cdot 0 = 0$ or $=\infty$? Or is this reasoning correct?
The functions converge pointwise to the function $0$ except at the origin. But the measure of the set $\{0\}$ equals to zero, so it does not matter what they converge to at the origin, so set it to $0.$ Then $f=0,$ but $\int |f_n|$ does not go to zero. Contradiction.