# Unsigned measurable function [on hold]

Let ff be an unsigned measurable function on (X,M,μ\mu). Prove that the function ν(E):=∫Efdμ\nu(E):=\int_E f d\mu is a measure on M, and that for all unsigned measurable gg, we have ∫gdν=∫gfdμ\int g d\nu=\int g f d\mu

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I assume that by “unsigned function” you mean “non-negative function”.

The set function ν\nu is a measure if the following hold: 1) ν\nu is non-negative; 2) ν(∅)=0\nu(\emptyset)=0; 3) ν(⋃n∈NAn)=∑n∈Nν(An)\nu(\bigcup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}\nu(A_n) for every countable family of pairwise disjoint sets {An}n∈N\{A_n\}_{n\in\mathbb{N}} (σ\sigma-additivity).

1) Since μ\mu is a measure and f≥0f\geq 0 (i.e. f(x)≥0f(x)\geq 0 a.e. x∈Xx\in X), ν\nu itself is non-negative: for all A∈MA\in\mathcal{M} we have ν(A)=∫Afdμ≥0\nu(A)=\int_Af\;d\mu \geq 0;

2) ν(∅)=0\nu(\emptyset)=0 is clear;

3) recall that μ\mu is a measure, therefore it is σ\sigma-additive. We can find a sequence of positive simple functions {sk}k\{s_k\}_k approximating ff from below s.t. ∫Efdμ=limk∫Eskdμ\int_Ef\;d\mu=\lim_k\int_Es_k\;d\mu (see the definition of the abstract integral). Moreover, if s=∑Mj=1αj1Bjs=\sum_{j=1}^M\alpha_j1_{B_j} is a simple function, then we define the integral ∫Esdμ=∑Mj=1αjμ(E∩Bj)\int_E s\;d\mu=\sum_{j=1}^M\alpha_j\mu\left(E\cap B_j\right). Please, let me drop the range in the finite sum: in what follows, when I will sum over jj, I will mean a finite sum. Thus, let 0\leq s_k=\sum_j\alpha_j^k 1_{B_j^k}\uparrow f0\leq s_k=\sum_j\alpha_j^k 1_{B_j^k}\uparrow f. Hence, by \sigma\sigma-additivity of \mu\mu:

\mu\left(\left(\bigcup_n A_n\right)\cap B_j^k\right)=\mu\left(\bigcup_n\left(A_n\cap B_j^k\right)\right)=\sum_n\mu\left(A_n\cap B_j^k\right),

\mu\left(\left(\bigcup_n A_n\right)\cap B_j^k\right)=\mu\left(\bigcup_n\left(A_n\cap B_j^k\right)\right)=\sum_n\mu\left(A_n\cap B_j^k\right),

which we use as follows:
\begin{align*}
\nu\left(\bigcup_n A_n\right) &=\int_{\bigcup_n A_n}f\; d\mu \\
&=\int 1_{\bigcup_n A_n}f\;d\mu \\
&=\lim_k\int 1_{\bigcup_n A_n}s_k\;d\mu \\
&=\lim_k\sum_j\alpha_j^k\mu\left(\left(\bigcup_n A_n\right)\cap B_j^k\right) \\ &=\lim_k\sum_j\alpha_j^k\sum_n\mu\left(A_n\cap B_j^k\right) \\ &=\sum_n\lim_k\sum_j\alpha_j^k\mu\left(A_n\cap B_j^k\right) \\
&=\sum_n\lim_k\int\sum_j\alpha_j^k 1_{A_n\cap B_j^k}\;d\mu \\
&=\sum_n\lim_k\int_{A_n} s_k\;d\mu \\
&=\sum_n\int_{A_n}f\;d\mu=\sum_n \nu(A_n)
\end{align*} \begin{align*}
\nu\left(\bigcup_n A_n\right) &=\int_{\bigcup_n A_n}f\; d\mu \\
&=\int 1_{\bigcup_n A_n}f\;d\mu \\
&=\lim_k\int 1_{\bigcup_n A_n}s_k\;d\mu \\
&=\lim_k\sum_j\alpha_j^k\mu\left(\left(\bigcup_n A_n\right)\cap B_j^k\right) \\ &=\lim_k\sum_j\alpha_j^k\sum_n\mu\left(A_n\cap B_j^k\right) \\ &=\sum_n\lim_k\sum_j\alpha_j^k\mu\left(A_n\cap B_j^k\right) \\
&=\sum_n\lim_k\int\sum_j\alpha_j^k 1_{A_n\cap B_j^k}\;d\mu \\
&=\sum_n\lim_k\int_{A_n} s_k\;d\mu \\
&=\sum_n\int_{A_n}f\;d\mu=\sum_n \nu(A_n)
\end{align*}
Please, notice that 1_{A\cap B}=1_A1_B1_{A\cap B}=1_A1_B, which we have used above.
I hope that there are no typos…

I broke out the spacing of the chain of equations. Feel free to rollback if this proves disagreeable, or (if not) study for future use.
– hardmath
yesterday

@hardmath: thanks, yesterday in the evening I was too lazy to write good TeX code… 🙂
– EM90
yesterday

Can you give me proof of the second part?
– indrasis
yesterday