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I am trying to prove this theorem:

Let ($\Omega$, $\mathfrak{F}$) be a measurable space. A mapping $\mu$: $\mathfrak{F}$ $\to$ $[0,\infty]$ is a measure iff

1 $\mu(\emptyset)=0$

2 $\mu(A \cup B)=\mu(A)+\mu(B) for A,B \in \mathfrak{F} \ and \ A \cap B = \emptyset$ (finite additivity)

3 If $A_n \subseteq A_{n+1} \ and \ \bigcup_{n=1}^{\infty}A_n = A, \ where \ A_n, A \in \mathfrak{F} \ then \ \lim_{n \to \infty} \mu(A_n) = \mu(A)$ (fundamental property of measure/continuity of measure)

Our definition of measure:

Let $\Omega$ be a nonempty set and F a $\sigma$-algebra on $\Omega$. We call ($\Omega$, $\mathfrak{F}$) a measurable space, and an element A $\in$ $\mathfrak{F}$.

A measure $\mu$ on ($\Omega$, $\mathfrak{F}$) is a mapping $\mu$: $\mathfrak{F}$ $\to$ $[0,\infty]$ satisfying:

i $\mu(\emptyset)=0$

ii $\forall$ pairwise disjoint sets $(A_n)_{n=1}^{\infty} \ in \ \mathfrak{F}, \ \mu(\bigcup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu(A_n)$ (countable additivity or $\sigma$-additivity)

So, the only if part was proved in class when we proved #3 for any measure space and clearly, ii implies #2

For the if part, #1 clearly follows from i. The main issue is to prove ii.

Before I give my proof attempt, here's what my professor says: we can replace #3 with either or both of the following if $\mu(\Omega) < \infty$:

4 If $A_{n+1} \subseteq A_n \ and \ \bigcap_{n=1}^{\infty}A_n = A, \ then \ \lim_{n \to \infty} \mu(A_n) = \mu(A)$

5 If $A_{n+1} \subseteq A_n \ and \ \bigcap_{n=1}^{\infty}A_n = \emptyset, \ then \ \lim_{n \to \infty} \mu(A_n) = \mu(\emptyset) = 0$.

How is that helpful?

Anyway, my proof attempt is the reverse of the proof of fundamental property of measure (I noticed that the proof relied on countable additivity):

Instead of converting increasing to pairwise disjoint, I convert pairwise disjoint to increasing.

Let $(A_n)_{n=1}^{\infty}$ be a pairwise disjoint collection in $\mathfrak{F}$.

Define $A \equiv \bigcup_{n=1}^{\infty} A_n$ and $B_n \equiv \bigcup_{k=1}^{n} A_k$.

Observe that $\bigcup_{n=1}^{\infty} B_n = \bigcup_{n=1}^{\infty} \bigcup_{k=1}^{n} A_k = \bigcup_{n=1}^{\infty} A_n = A, \ and \ B_n \subseteq B_{n+1}$.

By the fundamental property of measure, $\mu(A) = \lim_{n \to \infty} \mu(B_n)$

$=\lim_{n \to \infty} \mu(A_1 \cup A_2 \cup ... \cup A_n)$

$=\lim_{n \to \infty} \mu(A_1) + \mu(A_2) + ... + \mu(A_n)$ by finite additivity

$=\lim_{n \to \infty} \sum_{k=1}^{n} \mu(A_k)$

$=\sum_{k=1}^{\infty} \mu(A_k)$

$=\sum_{n=1}^{\infty} \mu(A_n)$

QED

If the above proof is correct, is @ShantDanielian's proof also correct? It seems the only difference is the defining of the increasing sets, though they can obviously be shown to be the same.

There seems to be something wrong philosophically...doesn't assuming continuity of measure mean that it is a measure already? Or should it be called continuity of measurable spaces...?

  • Update: It's actually not 'continuity of measure' since we don't know it's a measure. It's just like taking the conclusion of 'continuity of measure' the sets increase to another set then the 'measure', as in 'value assigned by $\mu$', of those sets increase to the 'measure' of that other set. I guess we may consider this conclusion as something like a property that a certain $\mu$ for a certain measurable space may satisfy.
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    $\begingroup$ First - your proof looks correct. Second, I think you can apply your professor's ideas to $A \setminus B_n$. $\endgroup$ Jul 2, 2014 at 4:32
  • $\begingroup$ @StephenMontgomery-Smith Thanks! I inadvertently understand limsup better :)) $\endgroup$
    – BCLC
    Jul 2, 2014 at 13:54

1 Answer 1

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1 up voted
First - your proof looks correct. Second, I think you can apply your professor's ideas to A∖Bn. – Stephen Montgomery-Smith 9 hours ago

@StephenMontgomery-Smith Thanks! I inadvertently understand limsup better :)) – BCLC 12 secs ago edit

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