I'm trying to prove a property of Lebesgue measure sets that says:

If the $A_{k}$'s are measurable and $A_{1} \supset A_{2} \supset A_{3} \supset \ldots,$ and if $\lambda (A_{1}) < \infty, $ then $$\lambda ( \bigcap_{k=1}^{\infty} A_{k} ) = \lim_{k \to \infty} \lambda (A_{k}).$$

And I'm trying to emulate the proof of the previous property that says:

If the $A_{k}$'s are measurable and $A_{1} \subset A_{2} \subset A_{3} \subset \ldots,$ then $$\lambda ( \bigcup_{k=1}^{\infty} A_{k} ) = \lim_{k \to \infty} \lambda (A_{k}).$$

And they express the union $\bigcup A_{k}$ as a disjoint union like this: $$\bigcup_{k=1}^{\infty} A_{k} = A_{1} \cup \bigcup_{k=2}^{\infty} (A_{k} \sim A_{k-1})$$ So I'm trying to create a increasing sequence from my decreasing one like this:

Say $(A_{k})= A_{1} \supset A_{2} \supset \ldots$ then $B_{k}=A_{1} - A_{k}$ and I'm not quite sure if $(B_{k})$ is an increasing sequence, and if so, I'm able to express this as a disjoint union like before?

I don't know if I'm on good track or I'm completely lost, can you help me with this please?

Also, what's the point of $\lambda (A_{1}) < \infty $ how can I use that?

Thank you guys!


You are definitely on the right track! Let's keep going from where you are:

We define $B_k := A_1 - A_k$. Then, as you can very easily prove, this is an increasing sequence of sets. It is now important to notice that $$\bigcup_{k = 1}^{\infty}B_k = \bigcup_{k = 1}^{\infty}(A_1 - A_k) = A_1 - \bigcap_{k = 1}^{\infty}A_k.$$ The only reason we are doing this is that we are now in the position to apply the previous result, which gives $$\lambda\Big(\bigcup_{k = 1}^{\infty}B_k\Big) = \lim_{k \to \infty}{\lambda(B_k)}.$$

Please notice that you can directly apply the result without proving it again, hence there is no need to make these sets into disjoint sets.

Let's write explicitly what those quantities are: $$\text{LHS} = \lambda\Big(\bigcup_{k = 1}^{\infty}B_k\Big) = \lambda\Big(A_1 - \bigcap_{k = 1}^{\infty}A_k\Big) = \lambda(A_1) - \lambda\Big(\bigcap_{k = 1}^{\infty}A_k\Big),$$ $$\text{RHS} = \lim_{k \to \infty}{\lambda(B_k)} = \lim_{k \to \infty}\lambda(A_1 - A_k) = \lambda(A_1) - \lim_{k \to \infty}\lambda(A_k).$$

Hence, $$\lambda(A_1) - \lambda\Big(\bigcap_{k = 1}^{\infty}A_k\Big) = \lambda(A_1) - \lim_{k \to \infty}\lambda(A_k),$$ which simplifies yielding the desired equality. It is a good exercise to convince yourself that you cannot conclude unless $\lambda(A_1) < \infty.$

($\infty - 1 = \infty = \infty - 2$, but $1$ and $2$ are different, so you can't cross out a quantity unless it is finite.)

It is worth mentioning that there is nothing special about $A_1$: the result continues to hold even if $\lambda(A_i) = \infty$ for $i = 1,\dots, N-1$ and $\lambda(A_{N}) < \infty.$ To prove this drop the first $N-1$ terms (they are irrelevant when computing the intersection) and replace $A_1$ with $A_N$ in the proof above.

There is a very easy example that shows the importance of the finiteness assumption: let $A_k = (k,\infty).$ Then $\lambda(A_k) = \infty$ for every $k$, but $\cap_k A_k = \emptyset$, so that $$0 = \lambda\Big(\bigcap_{k = 1}^{\infty}A_k\Big) \neq \lim_{k \to \infty}\lambda(A_k) = \infty.$$

  • 1
    $\begingroup$ That's great! Thak you soo much! I was actually crying about expressing $ \bigcup B_{k} $ as a disjoint union. And I'm already convinced that my $ (B_{k}) $ is an increasing sequence. Thank you again! $\endgroup$ – HelaHelheim Aug 25 '15 at 3:57
  • $\begingroup$ No problem, I am glad I could help! Also, I have edited the question several times trying to clean it from typos and I have tried to expand a little on the importance of the finiteness assumption. Make sure you check out this updated version :) $\endgroup$ – Giovanni Aug 25 '15 at 3:59
  • $\begingroup$ Actually, I was about to ask you for an example and you already did! That was really helpful, you help me understood that perfectly. $\endgroup$ – HelaHelheim Aug 25 '15 at 4:11

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