# 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.

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 $$\Omega$$

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

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

• You know that if $f_n$ is simple, then $\int_{\emptyset}f_nd\mu=0$, so you don't have to compute it again.
– Surb
Commented Dec 13, 2020 at 15:20
• True. Thanks a lot for your help, it's nice talking to someone who has a lot of knowledge in that topic. Did you gain that from books? If yes, are there any measure theory books you can recommend? Commented Dec 13, 2020 at 15:30
• You are welcome. I like very much the book : Real Analysis from Stein and Shakarchi. There is also a famous book on measure theory from Rudin (but I don't know it so well).
– Surb
Commented Dec 13, 2020 at 15:43

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.

• Thanks! So if I understood correctly it didn't work because I assumed $f$ to be simple. I'll try to repair the proof Commented Dec 13, 2020 at 14:31
• @Quotenbanane: No, it didn't work because you even didn't prove it for simple function (simple function are finite linear combinaison of characteristic function). When you write for example $\int fd\mu=\sum_{i=1}^\infty \alpha _i\mu(E_i)$, it's not clear if the RHS converges or not.
– Surb
Commented Dec 13, 2020 at 14:36
• I've edited my question and have added my new (hopefully correct) proof for 1). Could you give it a look? Commented Dec 13, 2020 at 15:16