measure of union of nested sets converges to measure of limiting set

[[concept]]

Topics

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Theorem (continuity of measure)

Suppose $\{E_k\}$ is a countable collection of measurable sets such that $E_1\subset E_2\subset \dots$ Then $m\left({\bigcup }_{k=1}^{\infty }{E}_k\right)={\lim }_{n\to \infty }m\left({\bigcup }_{k=1}^n\right)={\lim }_{n\to \infty }m(E_n)$

Proof

The second equality is because $E_n={\bigcup }_{k=1}^n{E}_k$. So it is enough to just show $m\left({\bigcup }_k{E}_k\right)={\lim }_{n\to \infty }m(E_n)$.

We can do this by showing the countable union as the countable disjoint union (recall that algebras have closure under finite disjoint countable unions).

So define $F_1=E_1$ and $F_k=E_k\setminus E_{k-1}$. Each $F_k$ is measurable since $F_k=E_k\cap E_{k-1}^c$ and the collection $\{F_k\}$ is disjoint. Then for all $n\in \mathbb{N}$, we have

\bigcup_{k=1}^n F_{k} = E_{n}\, \quad&\quad \bigcup_{k=1}^\infty F_{k} = \bigcup_{k=1}^\infty E_{k} \\ \implies m\left( \bigcup_{k=1}^\infty E_{k} \right) &= \sum_{k=1}^\infty m(F_{k}) \\ &= \lim_{ n \to \infty } \sum_{k=1}^n m(F_{k}) \\ &= \lim_{ n \to \infty } m\left( \bigcup_{k=1}^n F_{k} \right) \\ &= \lim_{ n \to \infty } m(E_{n}) \end{align}$$ Since [[Concept Wiki/measure of finite disjoint measurable sets is the sum of the measures]]. Thus we have shown the desired equality. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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