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Let $(X, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space with $\mu(X_n)<\infty$ and $X=\bigcup\limits_{n=1}^{\infty} X_n$. Let $$\mathcal{A}_n=\{E \in \mathcal{A} | E \subset X_n\}$$ and $\mu_n=\mu|_{\mathcal{A}_n}$ the restriction of $\mu$ onto $\mathcal{A}_n$. Now let $f_n:X_n \rightarrow [0,\infty)$ be a $X_n$-measurable function. If I extend $f_n$ to $X$ such that I set $f_n(x)=0$ for $x \in X_n^c$ and lets call this extension $\widetilde{f_n}: X \rightarrow [0,\infty)$, does then $$\int f_n d\mu_n = \int \widetilde{f_n} d\mu$$ hold? And how would I then proof this fact?

Thanks a lot in advnace!

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  • $\begingroup$ I assume you intended to write $\mu(X_n) < \infty$ in the first line? $\endgroup$ – Rhys Steele Jun 17 '17 at 7:48
  • $\begingroup$ Oh yes. I will correct it. Thanks. $\endgroup$ – vaoy Jun 17 '17 at 7:49
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    $\begingroup$ can you see that this is true if $f_n$ is a simple function? $\endgroup$ – Rhys Steele Jun 17 '17 at 7:57
  • $\begingroup$ I tried this approach, but unfortunatly not. $\endgroup$ – vaoy Jun 17 '17 at 7:59
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First consider the case that $f_n = \mathbb{1}_E$ for $E \in \mathcal{A}_n$. Since $\mathcal{A}_n \subset \mathcal{A}$, $E \in \mathcal{A}$ also and $\mu_n(E) = \mu(E)$ since $\mu_n$ is the restriction of $\mu$. Therefore $$\int \mathbb{1}_E d\mu = \int \widetilde{\mathbb{1}_E} d\mu_n$$

So by linearity your result holds if $f_n$ is simple. In general $$\int f d\nu = \sup \{\int \phi d\nu \mid \phi \mbox{ is simple and } \phi \leq f \}$$

But if $\phi$ is simple and $\phi \leq f_n$ then $\widetilde{\phi}$ is simple and $\widetilde{\phi} \leq \widetilde{f_n}$ and vice versa (it is clear that restricting an $\mathcal{A}$-simple function that is $0$ off of $X_n$ to $X_n$ gives an $\mathcal{A}_n$-simple function and that this restriction is the inverse of the map $f \mapsto \widetilde{f}$) so the set on the right hand side in the above is the same for both integrals of interest here and hence so are the integrals.

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