How can you prove that if $\mu$ and $\nu$ are finite measures and $n$ is a positive real number, then $\mu$+$n\nu$ is again a finite measure? Is that the same for $\sigma$-finite measures?
About sum of measures
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functional-analysis
measure-theory
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0Are you sure that $n$ could be any real number? – 2012-11-25
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0I'm confused: if you are given that $\mu (X) = K$ and $\nu (X) = K'$ then $\mu(X) + n \nu (X) = K + n K'< \infty$. What am I missing? – 2012-11-25
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0And you probably mean $n \in \mathbb R_{\ge 0}$. – 2012-11-25
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0@MattN. they could be signed measures, I suppose. [Wikipedia/ba space](https://en.wikipedia.org/wiki/Ba_space). – 2012-11-25
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2What is/are the axiom/axioms of a measure you have problems to check? – 2012-11-25
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0@kahen Okay. But how would that make any difference? If they are signed they still map $X$ to a real number and therefore the sum is also finite. – 2012-11-25