Let $f(x)$ be a nice function. Consider the integral $$\int_0^T ne^{n(x-T)}f(x)\ dx.$$
Apparently, this converges to $f(T)$ as $n\to\infty$. I can't see why. I've tried doing integration by parts, but it doesn't lead anywhere. I've put this in Wolfram Alpha for various choices of $f(x)$ and it seems to be true.
I can see that the factor $ne^{n(x-T)}$ seems to concentrate its mass around $x=T$ as $n\to\infty$, but then this would seem to suggest that the integral is almost $n\int_0^T \chi([2-\epsilon,2]) f(x)$. Not really sure what's going on!
Why does this integral converge to $f(T)$?
EDIT: I tried integration by parts as follows. Let $v=f(x)$ and $du=ne^{-n(x-T)}$. Then $$\int_0^T ne^{n(x-T)}f(x)\ dx = e^{n(x-T)}f(x)\Bigg|_0^T - \int_0^T e^{n(x-T)}f'(x)$$
$$=f(T) - e^{-nT}f(0)- \int_0^T e^{n(x-T)}f'(x),$$ which doesn't seem to lead anywhere.