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This is a theorem I'm trying to prove in Royden(4th ed).

Let $(X,\mathcal{M},\mu)$ be a measure space and $f$ be integrable over $X$.

  1. If $\{X_n\}_{n=1}^\infty$ is an ascending countable collection of measurable subsets of $X$ whose union is $X$, then $$ \int_X f~d\mu = \lim_{n\to \infty} \int_{X_n} f~d\mu.$$
  2. If $\{X_n\}_{n=1}^\infty$ is an descending countable collection of measurable subsets of $X$ then $$ \int_{\bigcap_{n=1}^{\infty}X} f~d\mu = \lim_{n\to \infty} \int_{X_n} f~d\mu.$$

The book says it follows from this theorem: Suppose $f$ is integrable on $X$, $\{X_n\}_{n=1}^\infty$, a disjoint countable colletction of measurable sets whose union is $X$. Then $$\int _X f~d\mu = \sum _{n=1}^\infty \int_{X_n} f~d\mu.$$ I however can't see it, and I ask for your help.

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2 Answers 2

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Hint for 1.: If $(X_n)_{n\geqslant1}$ is ascending with union $X$ and if every $X_n$ is measurable, then $Y_1=X_1$ and $Y_{n+1}=X_{n+1}\setminus X_{n}$ for every $n\geqslant1$ defines a sequence $(Y_n)_{n\geqslant1}$ of disjoint measurable subsets with union $X$.

Hint for 2.: If $(X_n)_{n\geqslant1}$ is descending with intersection $X$ and if every $X_n$ is measurable, then $Y_{n}=X_{1}\setminus X_{n+1}$ for every $n\geqslant1$ defines an ascending sequence $(Y_n)_{n\geqslant1}$ of measurable subsets with union $X_{1}\setminus X$. Furthermore, $f$ is integrable on $X_1$ hence $\int\limits_{X_1\setminus X}f\mathrm d\mu=\int\limits_{X_1}f\mathrm d\mu-\int\limits_{X}f\mathrm d\mu$ and, for every $n\geqslant1$, $\int\limits_{X_1\setminus X_{n+1}}f\mathrm d\mu=\int\limits_{X_1}f\mathrm d\mu-\int\limits_{X_{n+1}}f\mathrm d\mu$.

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Thanks${}{}{}{}{}$ –  Kuku Apr 5 '12 at 16:44
    
Thanks very much. For Hint 2, I guess I would have to use the result in 1. Is there another way of doing 2 without appealing to 1? –  Kuku Apr 5 '12 at 20:47
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We just "disjoint-ify" the ascending / descending collections of sets:

If $X_1\subseteq X_2\subseteq \cdots \subseteq X$ is an ascending (countable) sequence of sets with $$\bigcup_{n=1}^\infty X_n=X,$$ then let $Y_1=X_1$ and let $Y_n=X_n\setminus X_{n-1}$ for all $n>1$. Thus, for any $n$ we have that $$X_n=\bigcup_{i=1}^n Y_i,$$ and this is a disjoint union, so that $$\int_{X_n}f\,d\mu=\sum_{i=1}^n\int_{Y_{\,i}}f\,d\mu.$$ Because we now have a countable disjoint collection of sets $\{Y_i\}$ that have the property$$\bigcup_{n=1}^\infty Y_i=X,$$ by the cited theorem we know that $$\int_X f\,d\mu=\sum_{i=1}^\infty\int_{Y_i}f\,d\mu=\lim_{n\to\infty}\sum_{i=1}^n\int_{Y_i}f\,d\mu=\lim_{n\to\infty}\int_{X_n}f\,d\mu.$$

A similar argument will work for the descending case.

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Thanks. For the descending case I wrote $$X_1= (X_1\setminus X_2)\cup (X_2\setminus X_3)\cup \ldots \cup (\cap X_n),$$ but I got stuck. –  Kuku Apr 5 '12 at 16:46
    
Then I did $$\int_{X_1} f = \int_{X_n\setminus X_{n+1}} f + \int_{\cap X_n} f.$$ what to do next? –  Kuku Apr 5 '12 at 16:49
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