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Let $(X,\mathcal{A},\mu)$ be a measure space, let $A_1,A_2,A_3,\ldots\in \mathcal{A}$, and let $\sum_{j=1}^{\infty}\mu(A_j)<\infty$.

The task is to prove the following:

1) $\lim_{n\rightarrow \infty} \mu \Big( \bigcap_{j=1}^n A_j \Big)=0$

and

2) $\mu \Big( \bigcap_{j=1}^\infty A_j \Big)=0$

I don't see a fundamental difference between these two and I'm wondering if that means I've misunderstood something. My approach, in both cases, would be to argue that since the intersection is between an infinite amount (or approaching an infinite amount) of sets in the sigma algebra, that must mean that at some point a set will meet its complement, and therefore the intersection must be the empty set whose measure is 0.

Is this the right way to go about it? Or have I missed something? In what ways, if any, should the proofs differ?

Thanks in advance!

edit: I would be very grateful if your answers would contain explanations to why my initial approach can/can't be done, and not merely proofs of the above

Eugene Zhang
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user273860
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  • Since the sum of the measures converges, we must have that the $\mu(A_{n})$ must become really small for sufficiently large n's. Now taking intersections, the measure gets even smaller. We don't have any clue on what kinds of sets these abstract objects are, so we cannot say anything about empty intersection or so, other than that the measure must be very small. – TheOscillator Sep 24 '15 at 22:59
  • I don't think you should speak of "approaching" as part of the definition of $\bigcap_{j=1}^\infty A_j$. That set is defined by saying $x\in\bigcap_{j=1}^\infty A_j$ if and only if $\forall j\in{1,2,3,\ldots},\ x\in A_j$. ${}\qquad{}$ – Michael Hardy Sep 25 '15 at 03:13

1 Answers1

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1). Since $\sum_{j=1}^{\infty}\mu(A_j)<\infty$ and $\mu(A_j)\geqslant0$, we have $$ \lim_{n\to\infty} \mu(A_n)=0\quad\text{and }\quad \bigcap_{j=1}^n A_j \subset A_n $$ Thus $$ 0\leqslant\lim_{n\to\infty} \mu \Big( \bigcap_{j=1}^n A_j \Big)\leqslant\lim_{n\to\infty} \mu(A_n)=0 $$ which means $$ \lim_{n\to\infty} \mu \Big( \bigcap_{j=1}^n A_j \Big)=0 $$ 2). Since for any $n$ $$ \bigcap_{j=1}^\infty A_j \subset \bigcap_{j=1}^nA_j $$ Thus $$ 0\leqslant\mu \Big( \bigcap_{j=1}^{\infty} A_j \Big)\leqslant\lim_{n\to\infty} \mu(A_n)=0 $$ which means $$ \mu \Big( \bigcap_{j=1}^{\infty} A_j \Big)=0 $$

Edit: The infinite intersection of sets does not mean that at some point a set will meet its complement, and therefore the intersection must be the empty set whose measure is $0$. In this case, $\bigcap_{j=1}^{\infty} A_j $ may not be empty set but has measure $0$, which is known as null set.

Eugene Zhang
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  • Thanks for the answer, it's obvious to me now, that if the sum of the measure of all sets is finite, this value will be spread thin (to nothing) when considering an infinite amount of sets.

    But I'd like to know, is there no way one could argue that when considering the intersection of sets in an entire sigma algebra, the mere fact that the sigma algebra contains complements, means that an infinite intersection will always be the empty set? I'm extremely curious to know! Please tell me why my first approach will/won't work

     – user273860 Sep 24 '15 at 23:34
  • Let $a \ne \phi$ with $\mu (A)=0$ and let $A_j =A$ for every $j$.There is no reason to suppose that the common intersection of an infinite sequence must be empty. For another example, with the usual measure on the reals, let $A_j={0,j}$ for each positive integer $j$. – DanielWainfleet Sep 25 '15 at 04:55
  • Okay, I'm guessing this is under the assumption that the sigma algebra could be uncountable? If I had a sigma algebra with only finitely many sets, and I did an intersection on all of them, could I not be sure that this intersection would be empty? – user273860 Sep 25 '15 at 12:18