unions of measurable sets are measurable

[[concept]]

Topics

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Statements

Single Unions

Proposition

If $E_1,E_2$ are measurable, then $E_1\cup E_2$ are measurable

Proof

Let $A\subset \mathbb{R}$. Since $E_2$ is measurable, we know $m^{\ast }(A\cap E_1^c)=m^{\ast }(A\cap E_1^c\cap E_2)+m^{\ast }(A\cap \underset{(E_1\cup E_2{)}^c}{\underbrace{E_1^c\cap E_2^c}})$ Then we also know

A \cap(E_{1} \cup E_{2}) &= (A \cap E_{1}) \cup (A \cap E_{2}) \\ &= (A \cap E_{1}) \cup (A \cap E_{2} \cap E_{1}^c) \\ \end{align}

Taking outer measure we get

\implies m^*(A\cap(E_{1} \cup E_{2})) & \leq m^*(A \cap E_{1}) + m^*(A \cap E_{2} \cap E_{1}^c) \\ E_{1} \text{measurable } \implies&= m^*(A) - m^*(A \cap E_{1}^c) + m^*(A \cap E_{2} \cap E_{1}^c) \\ &= m^*(A) - m^*(A \cap (E_{1} \cup E_{2})^c) \end{align}$$ And this gives us the desired result.

Countable Unions

Theorem

If $E_1,\dots ,E_n$ are measurable, then ${\bigcup }_{k=1}^n{E}_k$ is measurable

Proof

by induction. $n=1$ is trivial. Suppose the statement holds for $n=k$. Then ${\bigcup }_{j=1}^{k+1}{E}_j=\left({\bigcup }_{j=1}^k{E}_j\right)\cup E_{k+1}$ And then since unions of measurable sets are measurable we get the desired result.

References

References

See Also

Mentions

Mentions

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