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Let $m$ be the Lebesgue Measure. If $\{A_k\}_{k=1}^{\infty}$ is an ascending collection of measurable sets, then

$$m\left(\cup_{k=1}^\infty A_k\right)=\lim_{k\to\infty}m(A_k).$$

Can someone share a story as to why this is called one of the "continuity" properties of measure?

Norbert
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David
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2 Answers2

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My understanding:

If a function $f$ is continuous and $\lim_{n\to \infty}x_n=x$, then $f(x)=f(\lim_{n\to \infty}x_n)=\lim_{n\to \infty}f(x_n)$. That is, you can take the limit out. Similarly, $\lim_{n\to \infty}\cup_{k=1}^nA_k=\cup_{k=1}^{\infty}A_k$, the "continuity" property of measure implies $$ m(\cup_{k=1}^{\infty}A_k)=m(\lim_{n\to \infty}\cup_{k=1}^nA_k)=\lim_{n\to \infty}m(\cup_{k=1}^nA_k)=\lim_{n\to \infty}A_n $$

Note $\{A_n\}_{n=1}^{\infty}$ is an ascending collection of measurable sets, so $A_n=\cup_{k=1}^nA_k$.

Coiacy
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Since $\{A_k\}_{k=1}^\infty$ is an ascending family of sets we can vaguely write that $$ \lim\limits_{k\to\infty} A_k=\bigcup\limits_{k=1}^\infty A_k \qquad(\color{red}{\text{note: this is not rigor!}}) $$ then this property can be written as $$ m\left(\lim\limits_{k\to\infty} A_k\right)=\lim\limits_{k\to\infty}m(A_k) $$ Which looks very similar to Heine definition of continuiuty.

Norbert
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    It can be also interpreted rigorously: for a sequence $A_k$ of sets we can define $\lim_{k\to\infty}A_k$ as $\limsup_kA_k:=\bigcap_k\bigcup_{n\ge k}A_n$ if this equals to $\liminf_kA_k:=\bigcup_k\bigcap_{n\ge k}A_n$, which is the case for ascending or descending sequences of sets. – Berci Jun 26 '13 at 23:38
  • I agree with Berci. There is a rigorous notion of limits of sets. Unfortunately, it seems like this fact is suppressed in most measure theory courses today. More annoyingly, professors continue to say "continuity of measure" even though they never explain what a "limit of sets" is... These same professors would be appalled if we defined continuity of functions without defining limits of sequences, but nonetheless this is the way it is. –  Jul 12 '18 at 16:12