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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a non-negative function. Define $\nu : M \rightarrow \mathbb{R}$, where $M$ is the Lebesgue σ-algebra of measurable functions, by $\nu(E) := \int_{E} f$. Prove that $\nu$ is countably additive.

for {$E_i$} disjoint sets, $\nu(\cup_i E_i) = \int_{\cup_i E_i} f = \sum_i \int_{E_i} f = \sum_i \nu(E_i)$

I need to show that $\int_{\cup_i E_i} f = \sum_i \int_{E_i} f $

But, how should i proceed further ?

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  • $\begingroup$ You only have to show the equality in the middle. And this follows from which integral convergence theorem? $\endgroup$ – Friedrich Philipp Mar 14 '16 at 8:24
  • $\begingroup$ Well, there are not too many standard integral convergence theorems... It's just an application of it and then you are done. Not much to prove here. $\endgroup$ – Friedrich Philipp Mar 14 '16 at 9:02
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The integral over a set $A$ is defined as

$$ \int_A f(x) dx := \int_\mathbb{R} 1_A(x) f(x) dx.$$

Since $$(\sum_{i=1}^\infty 1_{E_i})(x) = 1_{\cup_{i=1}^\infty E_i}(x)$$ holds, it only remains to show $$ \sum_{i=1}^\infty \int 1_{E_i}(x) \, f(x)\, dx = \int (\sum_{i=1}^\infty 1_{E_i})(x) \, f(x) \, dx .$$ But this follows from the monotic convergence theorem:

Choose $g_n (x):= f(x) \cdot \sum_{i=1}^n 1_{E_i}(x)$ and $g (x):= f(x) \cdot \sum_{i=1}^\infty 1_{E_i}(x)$, then $g(x)=\lim g_n(x)$ and $g_n$ is monoton and non-negative, thus

$$\lim_{n\to \infty} \int g_n(x) dx = \int g(x) dx.$$

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Let's first do this for a characteristic function $\chi_F$.

Let $E = \biguplus E_i$, where $\biguplus$ means disjoint union. Then, $$ \nu(E)= \int_A \chi_B = \int_{\biguplus A_i} \chi_B = \mu(B\ \cap \uplus A_i) \\= \mu(\biguplus (B \cap A_i)) = \sum \mu(B \cap A_i) = \sum \int_{A_i} \chi_B = \sum \nu(A_i). $$

Similarly the above proposition will follow for all simple functions. To prove it for a general function, we note that it is enough to take positive functions alone. So let $f >= 0 $, and let $f_n$ be a sequence of simple functions increasing to $f$, and without loss of generality,assume $f_n \geq 0$. Note that for simple functions, it is true that $\nu $ is countably additive. Now, I leave you to use monotone convergence theorem to finish your argument. I thought I should complete my argument, but that makes your job very easy, so I should leave something for you.

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