Is $\mathcal{M}(X)$ a Banach space?

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    2017
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SUMMARY

The normed space $\mathcal{M}(X)$, consisting of all complex regular Borel measures on a locally compact Hausdorff space $X$, is proven to be a Banach space under the total variation norm defined as $\|\mu\| := \lvert \mu\rvert (X)$. This conclusion is based on the completeness of the space, which is established by demonstrating that every Cauchy sequence in $\mathcal{M}(X)$ converges to a limit within the same space. The discussion emphasizes the importance of understanding the properties of locally compact Hausdorff spaces in relation to measure theory.

PREREQUISITES
  • Understanding of normed spaces and their properties
  • Familiarity with complex regular Borel measures
  • Knowledge of total variation norm
  • Concepts of Cauchy sequences and convergence in metric spaces
NEXT STEPS
  • Study the properties of locally compact Hausdorff spaces
  • Explore the concept of total variation in measure theory
  • Learn about the completeness of normed spaces
  • Investigate examples of Cauchy sequences in $\mathcal{M}(X)$
USEFUL FOR

Mathematicians, particularly those specializing in functional analysis, measure theory, and topology, will benefit from this discussion. It is also relevant for graduate students studying advanced mathematical concepts related to Banach spaces.

Euge
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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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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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