Behaviour of $L^1$ function at $\infty$ I have the following doubt..
If $f$ is $L^1(\mathbb{R}),$ then can we show $\lim\limits_{k \rightarrow \infty} \int\limits_k^\infty f(s)\,ds=0$
If so please outline the proof..

Let $f \in L^1(\mathbb{R})$. Then $\int_{\mathbb{R}} f(s) ds = I < \infty$ and furthermore $\forall k \in \mathbb{N}$:
$$\int_{\mathbb{R}} f(s) ds = \int f(s) 1_{(- \infty, k]} ~ ds + \int f(s) 1_{[k, \infty)} ~  ds$$
Now we have 
\begin{align*}
\lim_{k \rightarrow \infty} \vert \int_{k}^{\infty} f(s) ds \vert &\leq \lim_{k \rightarrow \infty} \int \vert f(s) \vert 1_{[k, \infty)} ~  ds \\ 
&= \int \lim_{k \rightarrow \infty} \vert f(s) \vert 1_{[k, \infty)} ~  ds \\
&= \int \lim_{k \rightarrow \infty} \vert f(s) \vert 0 ~  ds = 0
\end{align*}
where in the middle step we used dominated convergence via $\vert f(s)\vert  1_{(k, \infty)} \leq \vert f(s) \vert$.
