Is the Intersection of Sets with Measure 1 Also Measure 1?

[Chris L T521]

Here's this week's problem!

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Problem: Suppose that $A_n\subset[0,1]$ such that $m(A_n)=1$ for $n=1,2,3,\cdots$. Show that $\displaystyle m(\bigcap\limits_{n=1}^{\infty}A_n)=1.$

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[Chris L T521]

This week's problem was correctly answered by Ackbach. You can find his solution below.

[sp]Let $S=[0,1]$ be the universe of discourse, so we define set complements in terms of $S$: for any set $C\subseteq S$, $\overline{C}:= S \setminus C$. For each $j=1,2,3,\dots$ define $B_{j}=\overline{A_{j}}$. Since $m(S)=1$, it follows that $m(B_{j})=0$ for all $j=1,2,3,\dots$. Now
$$m \left( \overline{ \bigcap_{j=1}^{ \infty} A_{j}} \right)=m \left( \bigcup_{j=1}^{ \infty} \overline{A_{j}} \right) = m \left( \bigcup_{j=1}^{ \infty} B_{j} \right).$$
Now then, define the following new sequence of sets based on the $B_{j}$:
\begin{align*}
C_{1}&=B_{1} \\
C_{2}&= B_{2} \setminus B_{1} \\
C_{3}&= B_{3} \setminus (B_{1} \cup B_{2}) \\
C_{4}&= B_{4} \setminus (B_{1} \cup B_{2} \cup B_{3}) \\
\vdots \\
C_{n}&= B_{n} \setminus \left( \bigcup_{j=1}^{n-1}B_{j} \right).
\end{align*}

Then the elements of the sequence $\{C_{j}\}$ are mutually disjoint, and
$$\bigcup_{j=1}^{ \infty}C_{j}= \bigcup_{j=1}^{ \infty}B_{j},$$
and therefore
$$m \left( \bigcup_{j=1}^{ \infty}C_{j} \right)=m \left( \bigcup_{j=1}^{ \infty}B_{j} \right).$$
But $m(C_{j}) \le m(B_{j})$, since $C_{j} \subseteq B_{j}$. It follows that $m(C_{j})=0$. However, since the $\{C_{j}\}$ are mutually disjoint, it must be that
$$m \left( \bigcup_{j=1}^{ \infty}C_{j} \right) = \sum_{j=1}^{ \infty}m(C_{j})=0.$$
Since
$$m \left( \overline{ \bigcap_{j=1}^{ \infty} A_{j}} \right) =0,$$
it follows that
$$m \left( \bigcap_{j=1}^{ \infty} A_{j} \right)=1.$$
QED.[/sp]