# Problem of the Week #243 - Feb 07, 2017

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Consider the normed space $\mathcal{M}(X)$ of all complex regular Borel measures on a locally compact Hausdorff space $X$, with total variation norm $\|\mu\| := \lvert \mu\rvert (X)$, for all $\mu\in \mathcal{M}(X)$. Prove that $\mathcal{M}(X)$ is a Banach space.

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#### Euge

##### MHB Global Moderator
Staff member
Hi MHB community,

My apologies for the delayed post, I was unwell. No one answered this week's problem, but you can read my solution below.

Let $(\mu_j)$ be an absolutely summable sequence in $\mathcal{M}(X)$. We want to show $\sum_j \mu_j$ converges in $\mathcal{M}(X)$. Let $\lambda := \sum_j \lvert \mu_j\lvert$. Then $\mu_j <<\lambda$ for every $j$, so there exists a sequence $f_j\in \mathcal{L}^1(\lambda)$ such that $d\mu_j = f_j\, d\lambda$ for all $j$. Since $\sum_j \|f_j\|_1 = \sum_j \|\mu_j\| < \infty$ and $\mathcal{L}^1(\lambda)$ is complete, $\sum_{j = 1}^\infty f_j$ converges to a function $f\in L^1(\lambda)$. Define a measure $\mu$ on $X$ by the assignment $A \mapsto \int_A f\, d\mu$. It is an element of $\mathcal{M}(X)$, and the series $\sum_{j} \mu_j$ converges to $\mu$.

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