Using thm.13 on pg.374 in Royden and Fitzpatrick "Real analysis" fourth edition.

Question

I was trying to prove that $\nu$ is countably additive in the following question:

Let $f$ be a nonnegative function on a measure space $(X, \mathfrak{M}, \mu)$ and suppose that $f$ is integrable with respect to $\mu$. for each $E \in \mathfrak{M}$ define $$\nu(E) = \int_{E} f d\mu.$$ Prove that $\nu$ is a measure on $\mathfrak{M}.$

But in order to use thm.13 on pg.374 in Royden and Fitzpatrick "Real analysis" fourth edition, I must have a measure space $(X, \mathfrak{M}, \mu)$ but what I am trying to prove the countable additivity for is $E \in \mathfrak{M}$ and not $X$ itself, does this matters?

More precisely, my question is, if $(X, \mathfrak{M}, \mu)$ is a measure space, does this means that $(E, \mathfrak{M}, \mu)$ is a measure space? could anyone clarify this for me please?

Theorem 13 picture:

You do not prove countable additivity for any specific measurable set $E$. It is a property of the measure: you should prove that given any sequence of disjoint measurable sets $E_n\in\cal{M}$, then $\nu(\bigcup_nE_n)=\sum_n\nu(E_n)$. — GReyes, Jan 14 '20 at 07:44
And what is the relation between X and $\cup_{n} E_{n} = E$? @GReyes — Intuition, Jan 14 '20 at 07:47
You do not need any $E$. The $E_n$'s are subsets of $X$ that belong to your algebra $\cal{M}$. $E$ is only used as a generic argument to define $\nu$ through $f$ and the original measure $\mu$. — GReyes, Jan 14 '20 at 07:50
So I can use the theorem directly according to what you said .... am I correct?@GReyes — Intuition, Jan 14 '20 at 08:05

score 1 · Accepted Answer · answered Jan 14 '20 at 17:32

If you have $E=\bigcup_nE_n$ where $E_n\in\mathcal{M}$ you can use this result to prove countable additivity. First, the typical trick to express as an integral over the whole space, $$ \nu(E)=\int_Efd\mu=\int_Xf\chi_Ed\mu. $$ Next, it is clear that $X=E^c\cup E=E^c\cup\bigcup_nE_n$. Using the theorem, $$ \nu(E)=\int_{E^c}f\chi_Ed\mu+\sum_n\int_{E_n}f\chi_Ed\mu=0+\sum_n\int_{E_n}fd\mu=\sum_n\nu(E_n). $$ Since $f\ge 0$ $\mu$-a.e., $\nu(E)\ge 0$ for all $E\in\mathcal{M}$, so $\nu$ is a measure.

Using thm.13 on pg.374 in Royden and Fitzpatrick "Real analysis" fourth edition.

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