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Let $f:\mathbb R\rightarrow[0,\infty)$ be Lebesgue measurable. Is it true in general that the function:

$$\mu(E):=\int_E f$$

is a measure? The right hand side is the Lebesgue integral. I am having trouble proving the property of countable disjoint union. Suppose that $E_n$ ($n\geq1$) are disjoint sets, then:

$$\int_{\bigcup_{n\geq 1}E_n}f\quad\mathop{=}^{?}\quad\sum_{n\geq1}\int_{E_n}f$$

I know this equality is true for finite disjoint unions (using the linearity of the integral and the fact that $\chi_{\sum E_n}=\sum\chi_{E_n}$), but is it also true for countably infinite disjoint unions? (the linearity would fail in this case!).

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Hint: When you want to pass limits inside an integral what do you usually do? – Chris Janjigian Feb 28 '12 at 2:32
Hint: Use indicator functions and then justify the interchange of the integral and an appropriate sum. – cardinal Feb 28 '12 at 2:33
Thanks a lot to both! :) – wircho Feb 28 '12 at 2:49
up vote 6 down vote accepted

If $f$ is Lebesgue integrable, then yes, $\mu$ is an absolutely continuous measure with respect to Lebesgue measure.

Monotone Convergence of the indicator functions gives you the property of countable disjoint unions.

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Thanks a lot! I see it now! – wircho Feb 28 '12 at 2:48

There are two things you should look up:

  • Radon--Nikodym derivative. If $\nu(A) = \int_A f \; d\mu$ then $f = d\nu/d\mu$ is the Radon--Nikodym derivative of $\nu$ with respect to $\mu$. The Radon--Nikodym theorem says that under certain assumptions (including "absolute continuity" of $\nu$ with respect to $\mu$), a Radon--Nikodym derivative exists.
  • Density function. A Radon--Nikodym derivative is the same thing as a density function. Different ways of looking at it, maybe? Or maybe not.
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