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I am trying to prove that if $\{A_i\}$ is an increasing sequence of sets from some sigma algebra with $\mu(A_i) < \infty$ $\forall i$, where $\mu$ is some arbitrary measure, then $$\lim_{i\to\infty} \mu(A_i) = \mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$$

I would like to prove this using monotone convergence for sequences of real numbers, i.e. the theorem that the limit of an increasing sequence which is bounded above is its supremum. However, I am not sure if $\mu\bigg(\bigcup\limits_{i=1}^{\infty} A_i\bigg)$ is actually the supremum of $\{\mu(A_i)\}$. It is surely an upper bound, but I can't prove that it's the least upper bound.

If it is the supremum, how can I prove it?

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  • $\begingroup$ Maybe an easier approach. Consider the union of $B_i$ where $B_i=A_{i+1}\setminus A_i$, what can you say about the measure of it? $\endgroup$
    – Shashi
    Sep 27 '18 at 21:18
  • $\begingroup$ I have seen that approach in other answers on StackExchange and I agree that it's easier, but I would like to prove it using monotone convergence. $\endgroup$
    – Taliant
    Sep 27 '18 at 21:25
  • $\begingroup$ You cant use MCT without assuming the sets are measurable $\endgroup$
    – ibnAbu
    Sep 27 '18 at 21:47
  • $\begingroup$ I did. In my original post, I wrote "... is an increasing sequence of sets from some sigma algebra." Hence, they're measurable. $\endgroup$
    – Taliant
    Sep 27 '18 at 21:53
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Define $f_n:=\chi_{A_n}$ and $f:=\chi_{\cup_{i=1}^\infty A_i}$. Then since $(A_n)_{n\in\mathbb{N}}$ is an increasing sequence of sets, we have that $$0\le f_n\uparrow f.$$ Then, by monotone convergence theorem $$\mu(A_n)=\int f_n\operatorname{d}\mu\uparrow \int f\operatorname{d}\mu =\mu(\cup_{i=1}^\infty A_i).$$

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  • $\begingroup$ What if the sets are not measurable ? $\endgroup$
    – ibnAbu
    Sep 27 '18 at 21:49
  • $\begingroup$ I haven't yet learned the Beppo-Levi MCT. Nevertheless, this answer is helpful. I do have one question about the proof of it, and it relates to the main question I asked above. Namely, how can you show that $f$ is the supremum of $f_n$? It is clearly an upper bound, but how can one show that it is the least upper bound? $\endgroup$
    – Taliant
    Sep 27 '18 at 21:58
  • $\begingroup$ @taliant if x in in the union, then by definition of a union there exists $N\in\mathbb{N}$ such that $x\in A_N$ and so because the sequence of sets is increasing, for $n>N$ we have that $x\in A_n$ too, i.e. $f_n(x)=1=f(x)$ $\endgroup$
    – Bob
    Sep 28 '18 at 2:20
  • $\begingroup$ @ibnAbu if it is an objection, in the op hypothesis they are measurable. If it is a question, then $\mu(A_n)$ makes no sense at all... $\endgroup$
    – Bob
    Sep 28 '18 at 2:22
  • $\begingroup$ The proof of MCT for the integral uses the fact that needs to be proven, namely the one that OP wants to prove. I don't know if you can prove MCT for the integral without using it. What the OP want to use is MCT for sequences. But I might be wrong.... $\endgroup$
    – Shashi
    Sep 28 '18 at 6:35

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