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I'm studying about some qualities for sequences of measure functions,and I have this problem:

let $\mu$ be Dirac measure on the X= {$\frac{1}{n}; n\geq 1$} at the point $\frac{1}{i}$, I know that {$\mu_n$} is convergent pointwise, but I guess that the convergent measure is not a measure, Now my question is:

1- is my guess right?

2- How can I prove if {$\mu_n$} is a sequence of finite measures on the measurable space of (X,M) that is uniformly convergent to a finite measure $\mu$, then $\mu$ is a measure on $(X,M)$.

Any help would be great thanks.

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closed as unclear what you're asking by tomasz, астон вілла олоф мэллбэрг, user91500, Claude Leibovici, Davide Giraudo Jan 29 '17 at 10:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What do you mean by "convergent pointwise"? Weak-$*$ convergent in $C(X)^*$? What do you mean by "uniformly convergent"? $\endgroup$ – tomasz Jan 20 '17 at 7:43
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You can prove measure equality first for finite condition then use finiteness of m for monotonicity and continuity from below . Use the fact that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum . Now two inequality will prove easily .

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