convergence of probability measures?

amitgandhi_at_gmail.com
Date: 01/12/05 

Date: 12 Jan 2005 15:56:59 -0800

I have a question about convergence of probability measures that is
entering
some research I am doing. Here is the situation:

-Let us say our measurable space is an interval of the real line X =
[a, b], or a rectangle in R^n more generally.

-I have a sequence of probability measures \mu^t over X that converges
weakly to a limiting measure \mu. Each \mu^t is absolutely continuous
with respect to lebegue measure - i.e. they can be described by
densities. I have a belief as to what the limiting measure \mu is: \mu
is a delta mass over a special state z \in [a, b]. I want to prove
this.

-Here is the strategy I am trying to use for proof - and I wanted to
run it by anyone in this group. Instead of working with the sequence
\mu^t directly, I am going to deal with a disturbed sequence of
probability measures \m_{d}^t, where d is a number that represents a
disturbance parameter in the sense that for every t, as d->0, \m_{d}^t
converges weakly to \mu^t.

-For every fixed d > 0, the sequence (in t) of probability measures
\m_{d}^t converges weakly and strongly to a limiting probability
measure \m_{d}. Further I know that as d -> 0, these limiting measures
\m_{d} weakly converge to a probability measure \m, which I know is the
delta mass over z that was of original interest.

-From this, does it follow that I can show that \mu = \m, i.e., does
the limit \mu of the original sequence of probability measures \mu^t
equal \m, and hence the delta mass over z?

I was trying to see if I have the technical conditions needed to make
this result follow. Thanks to anyone in advanced.

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