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

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 $$ \int_\Omega f\,\mathrm d\mu=\int_0^\infty \mu(\{f\geq t\})\,\lambda(\mathrm dt). $$

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.