Assignment:
Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and $X: \Omega \rightarrow \bar{\mathbb{R}}$ a $\mathfrak{A}$-$\bar{\mathfrak{B}}$-measurable function.
Show that: $$ X \ {\text{is}}\ \mu{\text{-integrable}} \implies \sum_{k=1}^\infty\mu(\{\mid X\mid ≥ k\}) < \infty$$
My thoughts: Since $X$ is $\mu$-integrable we have $\int_{\Omega}Xd\mu <\infty$. Firstly I define $$M_k:=\{\omega\in\Omega: \mid X(\omega)\mid ≥k\})$$
Thus we also have $M_k \subset M_{k+1}$ I assume that $$\sum_{k=1}^\infty\mu(\{\mid X\mid ≥ k\}) = \sum_{k=1}^{\infty}\mu(M_k)= \sum_{k=1}^{\infty}\int\chi_{M_k}d\mu = \int\sum_{k=1}^{\infty}\chi_{M_k}d\mu = {\infty}$$
However $$\int\sum_{k=1}^{\infty}\chi_{M_k}d\mu = {\infty}\iff M:=\bigcap_{k=1}^{\infty}M_k \neq \emptyset $$ Since, if I suppose that $M = \emptyset$, I can find for arbitrary $\omega\in\Omega$ some $k_0\in\mathbb{N}$ such that for all $k≥k_0: \chi_{M_k}(\omega)=0$ so that
$$\int\sum_{k=1}^{\infty}\chi_{M_k}d\mu = \int\sum_{k=1}^{k_0}\chi_{M_k} + \sum_{k=k_0}^\infty\chi_{M_k} d\mu < \infty$$
And that is all I've got so far. Does this approach work? Or is there a better way to solve the problem? Because I'm struggling to find a contradiction.
Hint: Integrate the pointwise inequality $$\sum_{k=1}^\infty\mathbf 1_{\{\mid X\mid \geqslant k\}}=\lfloor |X|\rfloor\leqslant |X|.$$