measure of finite disjoint measurable sets is the sum of the measures

[[concept]]

Topics

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Theorem

Let $A\subset \mathbb{R}$ and let $E_1,\dots ,E_n$ be disjoint and measurable. Then $m^{\ast }\left(A\cap \left[{\bigcup }_{k=1}^n{E}_k\right]\right)={\sum }_{k=1}^n{m}^{\ast }(A\cap E_k)$

NOTE

if $E_1=E$ and $E_2=E^c$, this is the definition of measurability.

Proof

Via induction. Trivially true for $n=1$. Suppose that the equality is true for $n=m$. Now, suppose we have pairwise disjoint measurable sets $E_1,\dots ,E_{m+1}$ and $A\subset \mathbb{R}$. Since $E_k\cap E_{m+1}=\varnothing$ for all $k$, we have (since $E_{m+1}$ is measurable ) that

A \cap \left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E_{m+1} &= A \cap E_{m+1} \\ \text{and}\quad A \cap\left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E^c_{m+1} &= A \cap \left[ \bigcup_{k=1}^m E_{k} \right] \\ \implies m^*\left( A \cap \left[ \bigcup_{k=1}^{m+1}E_{k} \right] \right) &= m^*\left( A \cap \left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E_{m+1} \right) + m^*\left( A \cap\left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E^c_{m+1} \right) \\ &= m^*(A \cap E_{m+1} ) + m^*\left(A \cap \left[ \bigcup_{k=1}^m E_{k} \right] \right) \\ &= m^*(A \cap E_{m+1}) + \sum_{k=1}^m m^*(A \cap E_{k}) \end{align}$$ Where the last inequality is due to the induction hypothesis. Thus the result follows via induction. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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