# Lesbegue decomposition of sum of Borel measures.

Suppose $\mu_{n}$ is a sequence of positive Borel measures on $\mathbb{R}^{k}$and $$\mu(E)=\sum^{\infty}_{n=1}\mu_{n}(E)$$Assume $$\mu(\mathbb{R}^{n})<\infty$$Show that $\mu$ is a Borel measure. What is the relation between the Lesbegue decomposition of the $\mu_{n}$ and that of $\mu$? Prove that $$(D\mu)(x)=\sum^{\infty}_{n=1}(D\mu_{n})(x)$$almost everywhere.

It is not difficult to show $\mu$ is a Borel measure. It is also not difficult to show the Lesbegue decomposition of $\mu_{n}$ just add up to $\mu$, since the two parts are orthogonal to each other. However it is not clear to me why $D\mu$ is the sum of $D\mu_{n}$ almost everywhere, though I can deduce it from the theorem.

For example, I do not see how $D\mu$ could be different from $D\mu_{n}$ at a point if the Lesbegue Decomposition carries through, even though we have $D\mu_{n}$ defined in $L^{1}$ only. On the other hand, it is not difficult to image a sequence of measures defined on shrinking subsets of $\mathbb{R}^{k}$ such that $D\mu_{n}> 0$ for all $n$, but $\mu$ is a constant in some small set $K$ with positive Lesbegue measure. Then we would have $\sum D\mu_{n}>D\mu$ on $K$. This contradiction means there must be something wrong with my understanding.

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Where did Rudin ask that? – Jonas Meyer Jan 10 '13 at 3:04
@JonasMeyer: Real and Complex Analysis, Chapter on Differentiation, 1st Edition. Do not remember the problem number but should be 4 or 5. – Bombyx mori Jan 10 '13 at 5:14