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Suppose you have 2 probability spaces $(\Omega, \mu)$ and $(\Psi,\lambda)$. For every $t\in (0,1)$ let $f_t$ be a real-valued non-negative measurable function bounded by one, that is $f_t: \Psi\times\Omega \rightarrow [0,1]$, such that one has $$ \frac{1}{\mu(A)\lambda(B)}\int_{A \times B} f_t(x,w) ~d\mu(w) d\lambda(x) \xrightarrow[]{t\rightarrow 0} f~~~ $$ $$ \text{for every } A\times B \subset\Psi\times\Omega~\text{with nonzero measure}, $$ where $f$ is a fixed real number in $[0,1]$. The number $f$ does not depend on the choices of $A$ or $B$.

I would like to show that the function $x\mapsto \int_{\Omega} f_t (x,w)~d\mu(w)$ converges for $t\rightarrow 0$ almost surely to the constant function $f$. I have a strong feeling that this is true. However I don't know how to approach. If it makes things easier, I am especially interested in the case where $(\Omega,\mu) = (\Psi,\lambda) = ([0,1], \text{Lebesgue measure})$.

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    The condition *$f$ does not depend on the choices of "A" and "B"* looks strange. Did you mean to divide the integral by $\mu(A)\lambda(B)$?2012-06-03
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    If the sequence $(f_{t})$ is indexed by $t\in (0,1)$, how can you observe limiting when $t\to\infty$?2012-06-04
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    Sorry, of course I want to divide by $\mu(A)\lambda(B)$ and I am looking at $t\rightarrow 0$.2012-06-04

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