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Let $( \mathbb{R}^k , \mathcal{A} , m_{k} )$ be a Lebesgue measurable space, i.e., $m_{k}=m$ is a Lebesgue measure. Let $f: \mathbb{R^k} \to \mathbb{R}$ be a $m$-integrable function. Define a function $\mu : \mathcal{A} \to [0,\infty]$ by $$ \mu(A) := \int_{A} f(x) dx $$ with $A \in \mathcal{A}$. Then is $\mu$ a measure?

Working. When $A$ is an empty set, the integral is obviously 0. Now suppose $A_{0} , \dots , A_{n} \in \mathcal{A}$ as they are disjoint. Then $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) := \int_{\mathbb{R}^{k}} \chi_{\bigcup_{k=0}^{n} A_{k}}(x)f(x)dx $$ and we get $$ \chi_{\bigcup_{k=0}^{n} A_{k}} = \sum_{k=0}^{n} \chi_{A_k}$$ as $A_{k}$ are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking $n \to \infty$ we get the required result.

Question. Is there any error in the above working? And does it need any additional hypothesis to show $\mu$ is a measure? Or is it actually not a measure?

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What you did proves that $\mu$ is finitely additive. But it's not true that each finitely additive measure is $\sigma$-additive, so the passage "taking $n\to\infty$" need more details, like dominated convergence theorem. – Davide Giraudo Oct 8 '12 at 16:05
@DavideGiraudo Oh yes.. Thanks for that. I will note that. – julypraise Oct 8 '12 at 16:14

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