Let $f$ be Lebesgue Measurable and non-egative in $\mathbb{R}.$ If $E$ is measurable we define $$\phi(E) = \int_E f\,dx.$$ Then, $\phi$ is countably additive in the measurable sets.

I need $f$ to be non-negative and $f$ Lebesgue measurable so that the integral exists.

My try to prove the claim was using the proposition that says "for $A, B$ disjoint sets, then the integral of the union is the sum of the integrals of $f$ over each sets. And inductively, supposing it is valid for the union of $n-1$ disjoint sets:

Let $E = \bigcup_{n=i}^\infty E_i = (\bigcup_{n=1}^{n-1}E_i)\cup E_n,$ where $E_i\cap E_j = \emptyset $ whenever $i \neq j. $ Then

$$\int_E f\,dx = \int_{\bigcup_{n=1}^{n-1}E_i}f\,dx + \int_{E_n}f\,dx $$ $$ = \sum_{i=1}^{n-1} \int_{E_i} f\,dx + \int_{E_n} f\,dx = \sum_{i=1}^n \int_{E_i} f\,dx.$$

Never the less, i would like to verify if the proof is correct or if something is missing, which is possible, having revised To show a measure is countably additive .


  • 1
    $\begingroup$ Unfortunately you need argue more; you essentially rephrased finite additivity, and improving this to countable additivity requires certain limiting argument. But to be honest, I would recomment monotone convergence theorem (if available) for a single-line proof. $\endgroup$ – Sangchul Lee Oct 21 '17 at 0:28
  • $\begingroup$ Ok.:............. .. ..... $\endgroup$ – Trux Oct 21 '17 at 1:36

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