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Let $(X,\mathcal A,\mu)$ be a measure space. We define $\mu^{0}:\mathcal A\to [0,+\infty]$ by $\mu^{0}(A)=\sup\{\mu(B):B\subset A, \mu(B)<+\infty\}$. We want to show that $\mu^0$ is a measure on $(X,\mathcal A)$.

Clearly $\mu^0(\emptyset)=0$. Let $(A_n)$ be a sequence of pairwise disjoint members of $\mathcal A$. If $\mu(\bigcup\limits_{n=1}^{\infty}A_n)$ is finite, then $\mu^0(\bigcup\limits_{n=1}^{\infty}A_n)=\mu(\bigcup\limits_{n=1}^{\infty}A_n)=\sum\limits_{n=1}^{\infty}\mu(A_n)=\sum\limits_{n=1}^{\infty}\mu^0(A_n)$.

Let $\mu(\bigcup\limits_{n=1}^{\infty}A_n)=+\infty$. Let $\varepsilon>0$. Let $B\subset \bigcup\limits_{n=1}^{\infty}A_n$ such that $\mu^0(\bigcup\limits_{n=1}^{\infty}A_n)-\epsilon<\mu(B)$. Now $(A_n\cap B)$ is a sequence of pairwise disjoint members of $\mathcal A$ such that $\bigcup\limits_{n=1}^{\infty}(A_n\cap B)=B$ and so $\mu(B)=\sum\limits_{n=1}^{\infty}\mu(A_n\cap B)\leq \sum\limits_{n=1}^{\infty}\mu^0(A_n)$. Since $\varepsilon>0$ is arbitrary, therefore $\mu^0(\bigcup\limits_{n=1}^{\infty}A_n)\leq \sum\limits_{n=1}^{\infty}\mu^0(A_n)$. How to show the reverse inequality?

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Let $\{A_n\}$ be a disjoint sequence in $\mathcal A$, $A$ be the union of this sequence and $\epsilon >0$. By definition of $\mu^{0}(A_n)$ there exists $B_n \subset A_n$ such that $\mu(B_n)<\infty$ and $\mu^{0} (A_n) -\epsilon /2^{n} <\mu(B_n)$. If $n$ is a positive integer then $C_n=B_1 \cup B_2 \cup... \cup B_n$ is a subset of $A$ and $\mu (C_n) <\infty$. Hence $\mu^{0}(A) \geq \mu (C_n)=\sum_{i=1}^{n} \mu (B_i)$ because $B_i$'s are also disjoint. Hence $\mu^{0}(A) \geq \sum_{i=1}^{n} (\mu (A_i) -\frac {\epsilon} {2^{i}}) <\sum_{i=1}^{n} \mu (A_i) -\epsilon$. Since $n$ and $\epsilon$ are both arbitrary we get $\mu^{0}(A) \geq \sum_{i=1}^{\infty} \mu (A_i)$

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