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Let $\mu$ be a positive measure, $f$ be an extended $\mu$-integrable function. Define the set function $\nu$ on $\mathcal{M}$ by $$ \nu(E):=\int_E f~\text{d}\mu\qquad E\in \mathcal{M}.$$ I need assistance in showing that $\nu$ is signed measure.

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For countable additivity, try the dominated convergence theorem. –  Nate Eldredge Jan 25 '12 at 3:49
What part of the proof that $\nu$ is a signed measure trips you up? You have to show three things. What are they? –  tomcuchta Jan 25 '12 at 6:37
@tomcuchta: I have two of them done. I'm struggling with the countable additivity part... –  Bass Jan 25 '12 at 7:30

1 Answer 1

In order to add details to @Nate's comment, let $\{A_n\}\subset \mathcal M$ a sequence of pairwise disjoint sets. Put $B_n:=\bigcup_{j=1}^nA_k$ and $f_n(x):=\begin{cases}f(x)&\mbox{ if } x\in B_n\\\ 0&\mbox{otherwise}. \end{cases}$ If we fix $x\in X$, then either $x\in \bigcup_{n\in\mathbb N}A_n$, hence $x$ is in a set $A_{n_0}$ for some integer $n_0$, so $x\in B_n$ for $n\geq n_0$ and $f_n(x)=f(x)$ for $n\geq n_0$, or $x \notin \bigcup_{n\in\mathbb N}A_n$. In this case, $f_n(x)=0$ for all $n$. If we put $A:=\bigcup_{n\in\mathbb N}A_n$ then we got that for all $x\in X$: $\lim_{n\to\infty}f_n(x)=\mathbf 1_A(x)f(x)$. Since for all $n$ and for all $x$: $|f_n(x)|\leq |f(x)|$ and $f$ is integrable, we can apply dominated convergence theorem: $$\lim_{n\to \infty}\int_X f_n(x)d\mu=\int_X \lim_{n\to \infty}f_n(x)d\mu=\int_A f(x)d\mu=\nu(A)$$ and $$\int_X f_n(x)d\mu=\sum_{k=1}^n\int_{A_k}f(x)d\mu=\sum_{k=1}^n\nu(A_k),$$ and countable additivity follows.

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