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.


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

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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

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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