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working on a problem for measure theory and wanted to see if I'm on the right track.

I've shown that over a measurable space $(X,\mathcal M)$, $\sum_{j=1}^\infty a_j\mu_j$ where {$\mu_j$}$_{j=1}^\infty$ a sequence of measures, and {$a_j$}$_{j=1}^\infty \subset (0,\infty)$ is a measure.

Now I'm trying to find the conditions for which a specific example of this measure is finite, and $\sigma$-finite, the measure being $$\sum_{j=1}^\infty a_j\delta_{x_j}$$ where $\delta_x$ is the Dirac measure.

$\sum_{j=1}^\infty a_j\delta_{x_j}(X)$ = $\sum_{j=1}^\infty a_j$ since each $\delta_{x_j}$ = $1$ over the entire set X, and condition for finiteness of this measure should be the sum being finite. Is this correct?

Secondly, for $\sigma$-finiteness, my intuition tells me that X should be countable, since the Dirac measure is similar to the counting measure, but I'm not sure how to formally show conditions for finiteness or $\sigma$-finiteness.

Any help would be great! Thank you.

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The finiteness part is correct. The measure is always $\sigma$-finite. Each $x_j$ has finite measure (namely $a_j$) and $X = (\cup_j \{x_j\})\cup (X\setminus \cup_j \{x_j\})$ where $X\setminus \cup_j \{x_j\}$ is measurable with measure $0$.

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  • $\begingroup$ For the finiteness part, does it suffice to write what I have above to show that it's the condition for being finite? (i.e. $\sum_{j=1}^\infty a_j<\infty$). Also, could you elaborate a bit more on the second part - is X countable the only condition, or do I need $\sum_{j=1}^\infty a_j<infty$ as well? $\endgroup$ – Sank Sep 20 '18 at 3:43
  • $\begingroup$ Sorry, I needlessly added the $\sum a_j < \infty$ condition for sigma-finiteness. For the finiteness part, the point is that you want to show $\mu(A) < \infty$ for each measurable $A \subseteq X$. Since everything is monotone, showing that is equivalent to showing $\mu(X) < \infty$. But, as you noted, $\mu(X) = \sum_n a_n$. $\endgroup$ – mathworker21 Sep 20 '18 at 3:44
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    $\begingroup$ Yes. I mean that's the whole definition of $\sigma$-finite. If you wanted the condition to hold for all choices of $A_n$, then you can just take $A_1 = X$ and then you'll need $\mu$ to be a finite measure. $\endgroup$ – mathworker21 Sep 20 '18 at 4:08
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    $\begingroup$ Yes. See my previous comment. $\endgroup$ – mathworker21 Sep 20 '18 at 4:13
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    $\begingroup$ It does not affect the proof. $\endgroup$ – mathworker21 Sep 20 '18 at 4:40

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