Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!).

share|improve this question
1  
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
add comment

2 Answers 2

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.

share|improve this answer
    
Thanks a lot! I see it now! –  wircho Feb 28 '12 at 2:48
add comment

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.
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.