Show: $\sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f>k\right\})$ 

Show that für $f\colon (\Omega,\mathcal{A})\to\mathbb{R}_{\geq 0}$ measurable, it is
    $$
\sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f>k\right\}).
$$


I do not know if one needs it here, but we defined the integral of a non-negative measurable function $f\colon (\Omega,\mathcal{A})\to (\overline{\mathbb{R}},\overline{\mathcal{B}})$ by
$$
\int_{\Omega}f\, d\mu=\sup\left\{\sum_{k=1}^{n}(\inf\left\{f(x): x\in A_k\right\})\mu(A_k): n\in\mathbb{N}, A_1,\ldots,A_n\in\mathcal{A}\mbox{ disjoint },\Omega=\bigcup_{k=1}^{n}A_k\right\}.
$$

My first idea was to use this definition, but I did not come along with it...
Maybe you can help me to show that?
(I think, maybe it has something to do with monotone convergence theorem or majorated convergence?)
 A: Hint: Show the pointwise identity
$$
\sum_{k=1}^\infty \mu(\{f\geq k\})\mathbb{1}_{(k-1,k]}(t)\leq \mu(\{f\geq t\})\leq \sum_{k=1}^\infty \mu(\{f\geq k-1\})\mathbb{1}_{(k-1,k]}(t),
$$
for $t> 0$. Then integrate all three terms with respect to $\lambda$ (being the Lebesgue-measure). You will have to use monotone convergence theorem as well as the fact that for non-negative functions $f$ one has
$$
\int_\Omega f\,\mathrm d\mu=\int_0^\infty \mu(\{f\geq t\})\,\lambda(\mathrm dt).
$$
A: Funny, I have to prove the same statement!
I did it this way:
At first I show that it is pointwise
$$
\sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}\stackrel{\mathrm{(*)}}\leq f\stackrel{\mathrm{(**)}}\leq\sum\limits_{k=0}^{\infty}1_{\left\{f>k\right\}}.
$$
Proof of $(*)$:
Consider any $\omega\in\Omega$.
Case 1: $0\leq f(\omega)<1$. Then $1_{\left\{f\geq k\right\}}(\omega)=0~\forall~k\geq 1$, i.e. $\sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}(\omega)=0$ and therefore $f(\omega)\geq \sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}(\omega)$.
Case 2: $1\leq f(\omega)<\infty$, i.e. $f(\omega)=M$ with $1\leq M<\infty$. Then it is $1_{\left\{f\geq k\right\}}(\omega)=1$ for $1\leq k\leq\lfloor M\rfloor$ and therefore $\sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}(\omega)=\lfloor M\rfloor\leq M=f(\omega)$.
Case 3: $f(\omega)=\infty$. Then it is $1_{\left\{ f\geq k\right\}}(\omega)=1~\forall~k\geq 1$, i.e. $\sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}(\omega)=\infty=f(\omega)$.
The proof of $(**)$ is very similar, I leave it out here thefrefore.
Because of the monotony of the Lebesgue integral it is
$$
\int \sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}}\, d\mu\leq\int f\, d\mu\leq\int \sum\limits_{k=0}^{\infty}1_{\left\{f>k\right\}}\, d\mu.
$$
Now I defined the fundamental functions
$$
f_n:=\sum\limits_{k=1}^{n}1_{\left\{f\geq k\right\}},~~g_n:=\sum\limits_{k=0}^{n}1_{\left\{f>k\right\}}
$$
which are non-negative and measurable. Moreover it is
$$
f_n\uparrow \sum\limits_{k=1}^{\infty}1_{\left\{f\geq k\right\}},~~g_n\uparrow \sum_{k=0}^{\infty}1_{\left\{f>k\right\}}.
$$
Using the theorem of monoton convergence and the definition of the Lebesgue integral for non-negative and measurable fundamental functions, it follows
$$
\int \sum_{k=1}^{\infty}1_{\left\{f\geq k\right\}}\, d\mu=\lim_{n\to\infty}\int\sum_{k=1}^{n}1_{\left\{f\geq k\right\}}\, d\mu=\lim\limits_{n\to\infty}\sum_{k=1}^{n}\mu(\left\{f\geq k\right\})=\sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})
$$
resp.
$$
\int\sum\limits_{k=0}^{\infty}1_{\left\{f>k\right\}}\, d\mu=\lim\limits_{n\to\infty}\int\sum\limits_{k=0}^{n}1_{\left\{f>k\right\}}\, d\mu=\lim\limits_{n\to\infty}\sum\limits_{k=0}^{n}\mu(\left\{f>k\right\})=\sum\limits_{k=0}^{\infty}\mu(\left\{f>k\right\}).
$$
So that's it.
I am excited to hear your opinions to my proof.
