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I am trying to understand the following Corollary to the Dominated Convergence Theorem. Suppose, for some $t_0 \in [a,b] $, we have that

$ \lim_{t \to t_0} f(x,t) = f(x,t_0)$,

for each $x \in X$, and that there exists an integrable function $g$ on $X$ such that $|f(x,t)|\leq g(x)$, then $\int f(x,t_0)d\mu (x) = \lim_{t \to t_0} \int f(x,t)d\mu (x)$

The argument goes as follows:

Let $(t_n)$ be a sequence in $[a,b]$ which converges to $t_0$ and apply the Dominated convergence theorem to the sequence $(f_n)$ defined by $f_n(x)=f(x,t_n)$

The following is where I have been stuck.

We have, by the Domiated convergence theorem, that $ \int f(x,t_0)d\mu (x) = \lim_{t_n \to t_0} \int f(x,t_n)d\mu (x)$

My question is, I don't see why $ \lim_{t_n \to t_0} \int f(x,t_n)d\mu (x) = \lim_{t \to t_0} \int f(x,t)d\mu (x)$

I would be very grateful for any explanation of this.

Is this by the sequential characterization of limits, i.e we have for every sequence, $(t_n) \to t_0$ we have that

$\lim_{t_n \to t_0} \int f(x,t_n)d\mu (x) = \int f(x,t_0)d\mu (x)$ form which it follows that $\lim_{t \to t_0} \int f(x,t)d\mu (x) = \int f(x,t_0)d\mu (x)$

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I think your interpretation is correct. –  Dylan Moreland Sep 30 '11 at 1:07
    
Thanks for the clarification –  Shibi Vasudevan Sep 30 '11 at 1:41

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