Doubt in Proof of lower bound in Cramer's theorem I read proof of Cramer's theorem from the book : 'Large Deviations and Applications' by A.Dembo and Ofer Zeitouni.
I have a doubt in the proof of the lower bound. 

Cramer's theorem(Lower bound) : Let $ X_1, X_2 , \ldots $ be i.i.d random variables with law $\mu$ . Let $S_n = \sum_{i = 1} ^{n} X_i $ , and let $\mu_n$ be the law of $S_n / n$. Let $\Lambda$ be the cumulant generating function of $X_1$ , and let $\Lambda^{*}$ be the corresponding rate function.We then have
  $$ \liminf\limits_{n \rightarrow \infty } \mu_n((-\delta,\delta)) \geq  \inf_{\lambda \in \mathbb{R}} \Lambda(\lambda)  = -\Lambda^{*}(0)$$

I have read the proof of the above under the assumptions that $\mu$ has bounded support , and would  like to prove it without this assumption.
Without any assumptions on the support of $\mu$ ,I have been able to prove that for all $M$ sufficiently large
$$ \liminf\limits_{n \rightarrow \infty } \mu_n((-\delta,\delta)) \geq  \inf_{\lambda \in \mathbb{R}} \Lambda_M(\lambda)$$
where 
$ \Lambda_M(\lambda) = log \int_{-M}^{M}  exp(\lambda x) d\mu(x)$

TO SHOW :$$ \liminf\limits_{n \rightarrow \infty } \mu_n((-\delta,\delta)) \geq  \inf_{\lambda \in \mathbb{R}} \Lambda(\lambda)$$

We have that $\inf_{\lambda \in \mathbb{R}} \Lambda_M(\lambda)$ increases( in $M$ ) , but I do not know if it increases to $\inf_{\lambda \in \mathbb{R}} \Lambda(\lambda)$.
 A: A few things here. First of all, you have stated the lower bound of Cramer's theorem incorrectly. It should read
$$\liminf_{n\to\infty}\frac1n\log\mu_n(U)\ge -\inf_{x\in U}I(x)$$
for any $U\subset\mathbb R$, where $I=\Lambda^*$. Since $x\in U$ implies $U\supseteq(x-\delta,x+\delta)$ for some $\delta>0$, this is equivalent to showing
$$\liminf_{n\to\infty}\frac1n\log\mu_n((x-\delta,x+\delta))\ge-I(x).$$
In fact, as you have done (and as DZ do in their book) you may assume $x=0$. However, this does not make the proof substantially easier, and I believe it makes it less obvious precisely what we are doing at certain points. So I will keep $x$ general. Note also that there is no need to assume $\mu$ has bounded support in what follows.
Let $\lambda^*$ be the maximizer of $\lambda x-\Lambda(\lambda)$, i.e. $I(x)=\lambda^*x-\Lambda(\lambda^*)$. Note that $\lambda^*$ is characterized by $\lambda^*=I'(x)$ and $x=\Lambda'(\lambda^*)$. We have $\Lambda'(\lambda)=\frac{M'(\lambda)}{M(\lambda)}$, where $M(\lambda)=\int e^{\lambda z}\mu(dz)=e^{\Lambda(\lambda)}$. In particular we have $x=\Lambda'(\lambda^*)=e^{-\Lambda(\lambda^*)}\int ze^{\lambda^* z}\mu(dz)$. Now define the measure $\nu$ by
$$
\frac{d\nu}{d\mu}(z) = \exp\big(\lambda^*z - \Lambda(\lambda^*)\big).
$$
Observe that $\int x\nu(dz)=\int ze^{\lambda^*z-\Lambda(\lambda^*)}\mu(dz)=x$. Hence, if $\nu_n$ is the law of $S_n/n$ when $\{X_i\}$ is iid with common law $\nu$, it follows by the law of large numbers that $\nu_n((x-\varepsilon,x+\varepsilon))\to1$ for any $\varepsilon>0$. Fixing $0<\varepsilon<\delta$, we find
\begin{align*}
\mu_n((x-\delta,x+\delta))&\ge\mu_n((x-\varepsilon,x+\varepsilon))=\int_{|\sum z_i-nx|<n\varepsilon}\mu(dz_1)\ldots\mu(dz_n)\\
&\ge e^{-n\lambda^*(x+\varepsilon)}\int_{|\sum z_i-nx|<n\varepsilon}e^{\lambda^*\sum z_i}\mu(dz_1)\ldots\mu(dz_n)\\
&=e^{-n[\lambda^*(x+\varepsilon)-\Lambda(\lambda^*)]}\int_{|\sum z_i-nx|<n\varepsilon}\nu(dz_1)\ldots\nu(dz_n)\\
&=e^{-n[I(x)+\lambda^*\varepsilon]}\nu_n((x-\varepsilon,x+\varepsilon)).
\end{align*}
Hence, letting $n\to\infty$, we see
$$\liminf_{n\to\infty}\frac1n\log\mu_n((x-\delta,x+\delta))\ge-[I(x)+\lambda^*\varepsilon].$$
The result follows by letting $\varepsilon\to0$.
