Show that h is lower semicontinuous The question is as follows:
Let C_n be a sequence of probability measures on a metric space $S$, and $f:S\rightarrow[0,\infty]$ a function with the property that for all $x\in S$ we have
$\lim_{\epsilon \downarrow 0} \liminf_{n\rightarrow \infty} \frac{1}{n} \log C_n(B(x,\epsilon))=\lim_{\epsilon \downarrow 0} \limsup_{n\rightarrow \infty} \frac{1}{n} \log C_n(B(x,\epsilon))=-h(x).$
$B(x,\epsilon)$ is the ball of radius $\epsilon$ aound $x$. Show that h is lower semicontinuous.
I have tried using the Large Deviation Principle, but I run into the problem, that the rate function I(x) is assumed to be LSC. And hence I can not use this to show that h(x) is LSC. Any ideas?
 A: Assume on one point $\hat{x}$ that $h$ is not lower-semicontinuous, then there exists a $\delta > 0$ such that for any $\epsilon > 0$, there exists $x\in B(\hat{x},\epsilon)$ such that $h(x) < h(\hat{x})- \delta$. $\ \ \ \ \ \ \ $(*)
Also because of the assumption you made ( the limit equation), for the same $\delta > 0$,  there exists a $\epsilon^* > 0$ such that
\begin{equation}
\limsup_{n\rightarrow\infty}\frac{1}{n}\log C_n(B(\hat{x}, \epsilon^*)) < -h(\hat{x}) + \frac{\delta}{2}
\end{equation}
which implies the probability $C_n(B(\hat{x}, \epsilon^*))$ is upper bounded by $e^{-n(h(\hat{x}) - \frac{\delta}{2})}$ for $n$ large.
Due to (*), we could find a $x^*\in B(\hat{x}, \epsilon^*)$, such that $h(x^*) < h(\hat{x})- \delta$, and due to the assumption again, we could find a ball around $x^*$ such that $B(x^*, r)\subset B(\hat{x}, \epsilon^*)$ and
\begin{equation}
\liminf_{n\rightarrow\infty}\frac{1}{n}\log C_n(B(x^*, r)) > -h(x^*) -  \frac{\delta}{4} > -h(\hat{x}) + \frac{3\delta}{4}
\end{equation}
which implies the probability $C_n(B(x^*, r))$ is lower bounded by $e^{-n(h(\hat{x}) - \frac{3\delta}{4})}$ for $n$ large.
We know $C_n(B(x^*, r))\leq C_n(B(\hat{x}, \epsilon^*))$ holds for each $n$ by the set inclusion, but when $n$ large,  the former's lower bound $e^{-n(h(\hat{x}) - \frac{3\delta}{4})}$ is strictly larger the later's upper bound $e^{-n(h(\hat{x}) - \frac{\delta}{2})}$. This yields a contradiction.
