Is $\nu(E) := \int_E f \, d\mu$ a measure on $\Omega$ Let $(\Omega, \mathcal{A},\mu)$ be a measure space. and $f:\Omega \rightarrow \bar{\mathbb{R}}^+_0$ a measurable function.
Show that $\nu(E) := \int_E f \, d\mu$ is a measure on $\Omega$

I thought about it like that...
A measure must satisfy 2 conditions.

*

*$\nu(\emptyset)=0$

*$\nu(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} \nu(A_i)$ with $A_i\in\mathcal{A}$ disjoint.

The first condition looks pretty easy. Since $f$ is measurable it can be described by a step function $f = \sum_{k=1}^{\infty} \alpha_k \mathcal{X}_{E_k}$ (with $\mathcal{X}$ being the indicator function and $\alpha_i$ being the function value in that point)
Furthermore, $\int_E f \, d\mu= \sum_{k=1}^{\infty} \alpha_k \mu(E)$ with $E\in \mathcal{A}$
$$\nu(\emptyset) = \int_\emptyset f\, d\mu=\sum_{k=1}^{\infty} \alpha_k \mu(\emptyset)=\sum_{k=1}^{\infty} \alpha_k * 0=0$$
So 1) seems true to me.
I've thought about the second condition too, but I'm not sure about it.
$$\nu(\bigcup_{i=1}^{\infty}A_k)=\int_{\bigcup_{i=1}^{\infty}A_k} f\, d\mu = \sum_{k=1}^{\infty} \alpha_k \mu(\bigcup_{i=1}^{\infty}A_k)=\sum_{k=1}^{\infty} \alpha_k \sum_{i=1}^{\infty} \mu(A_i) $$
$$=\sum_{k=1}^{\infty} \sum_{i=1}^{\infty}  \alpha_k\mu(A_i) =\sum_{k=1}^{\infty} \int_{A_i}  f\,d\mu = \sum_{k=1}^{\infty} \nu(A_i) $$
So 2) must be true too and 1)+2) $\Rightarrow$ that $\nu$ is a measure on $\Omega$
Have I made a mistake? Hoping for helpful inputs, thanks in advance =)

EDIT:
For the 1st:
Since $f$ is positive and measurable, there exists a sequence of simple functions $f_n(x) \nearrow f(x)$ for all $x$.
Because $f_n$ simple $\Rightarrow f_n(x)= \sum_{k=1}^n\alpha_i\mathcal{1}$. Let $(f_n)_{n\in\mathbb{N}}$ be the monotone function sequence, consisting of simple functions with $\lim_{n\rightarrow \infty} f_n = f$
$$\nu(\emptyset) = \int_\emptyset f\, d\mu=\int_\emptyset \lim_{n\rightarrow \infty}f_n\, d\mu\overset{MCT}{=}\lim_{n\rightarrow \infty}\int_\emptyset f_n\, d\mu=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \alpha_k \mu(\emptyset\cap E_k)=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} 0 = 0$$
 A: Hint
Both doesn't seem correct.
For the first one

*

*If $f$ is simple, then $$f(x)=\sum_{k=1}^n \alpha _k\boldsymbol 1_{E_k},$$
(the sum is finite). Then $$\mu(\emptyset)=\sum_{k=1}^n\alpha _k\mu(\emptyset\cap E_k)=\sum_{k=1}^n\alpha _k\mu(\emptyset)=0.$$


*If $f$ is positive, there is a sequence of simple function s.t. $f_n(x)\nearrow f(x)$ for all $x$, then using Monotone convergence theorem will do the work.


*If $f$ is measurable, then $f=f^+-f^-$ where $f^+(x)=\max\{f(x),0\}$ and $f^-(x)=-\min\{f(x),0\}$ are both positive, and try to apply the previous step.
For the second one
Still doesn't work. Let $\{A_i\}_{i\in\mathbb N}$ disjoints and measurable. If $f$ is simple, i.e. $$f(x)=\sum_{i=1}^n\alpha _i\boldsymbol 1_{E_i},$$
where $E_i$ are measurable, then
\begin{align*}
\int_{\bigcup_{j=1}^\infty A_i}f\,\mathrm d \mu=\sum_{i=1}^n\alpha _i\mu\left(E_i\cap \bigcup_{j=1}^\infty A_i\right)&=\sum_{i=1}^n\alpha _i\mu\left( \bigcup_{j=1}^\infty E_j\cap A_i\right)\\
&=\lim_{m\to \infty }\sum_{i=1}^n\alpha _i\mu\left(\bigcup_{j=1}^mE_i\cap A_j\right).
\end{align*}
I let you conclude from here.
When $f\geq 0$ and $f$ measurable, apply the same recipe than the first one.
