Is $\nu(E) := \int_E f \, d\mu$ a measure on $\Omega$

Question

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 $\Omega$

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I thought about it like that...

A measure must satisfy 2 conditions.

1. $\nu(\emptyset)=0$
2. $\nu(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} \nu(A_i)$ 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 =)

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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 $\lim_{n\rightarrow \infty} f_n = f$

$$\nu(\emptyset) = \int_\emptyset f\, d\mu=\int_\emptyset \lim_{n\rightarrow \infty}f_n\, d\mu\overset{MCT}{=}\lim_{n\rightarrow \infty}\int_\emptyset f_n\, d\mu=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \alpha_k \mu(\emptyset\cap E_k)=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} 0 = 0$$

Answer 1

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.