Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $(X\_n:\Omega\rightarrow \mathbb{R}^m)\_n$ be a sequence of i.i.d. random variables and let $L:\mathbb{R}^m\rightarrow [0,\infty)$ be Lipschitz. Let $\mu\_n:=\frac1{n} \sum\_{k=1}^n \delta\_{X\_k}$. Are there conditions under which:
$$
\mathbb{P}\left(|\mathbb{E}\_{X\sim\mu\_n}[L(X)]-\mathbb{E}\_{X\sim Law(X\_1)}[L(X)]|\geq t\right)\leq \exp\left(
I(t)
\right),
$$
where $I$ is a [good rate function](https://en.wikipedia.org/wiki/Rate_function)?



Concentration Inequality for Bounding Lipschitz Empirical Lass
Your inequality is trivial and useless as written. On its left-hand side we have a probability which is $\le1$ and goes to $0$ as $t\to\infty$, whereas on the right-hand side we have an expression which is $\ge1$ and goes to $\infty$ as $t\to\infty$, because for any good rate function $I$ on $[0,\infty)$ we have $I(t)\to\infty$ as $t\to\infty$. Also, the left-hand side of your inequality depends on $n$, whereas the right-hand side does not. Overall, this is not how a good rate function is used.


The corrected version of your inequality is as follows:
\begin{equation\*}
 P(|T\_n|\ge t)\le e^{-nI(t)} \tag{1}
\end{equation\*}
for some good rate function $I$ and all real $t\ge0$, where
\begin{equation\*}
 T\_n:=\frac1n\,\sum\_1^n Y\_k,\quad Y\_k:=L(X\_k)-EL(X\_k). 
\end{equation\*}


Moreover, the conditions that the $X\_k$'s take values in $\mathbb R^m$ and that $L$ is Lipschitz are of no relevance. Instead, all we need here is that the $Y\_k$'s are iid zero-mean real-valued random variables.


Once all this preliminary cleaning is done, we can now say that for (1) to hold (for some good rate function $I$ and all real $t\ge0$), it is sufficient that
\begin{equation\*}
 Ee^{h|Y\_1|}<\infty \tag{2}
\end{equation\*}
for some real $h>0$. (It is also easy to see that (2) is also necessary for (1).)


Indeed, assume (2) holds for some real $h>0$. By Markov's inequality, for any $t\ge0$,
\begin{equation\*}
 P(T\_n\ge t)\le e^{-ntx+nl(x)}
\end{equation\*}
for all $x\ge0$, where
\begin{equation\*}
 l(x):=\ln Ee^{xY\_1}
\end{equation\*}
and hence
\begin{equation\*}
 P(T\_n\ge t)\le e^{-nI\_+(t)},
\end{equation\*}
where
\begin{equation\*}
 I\_+(t):=\sup\_{x\ge0}(tx-l(x)). 
\end{equation\*}
Similarly,
\begin{equation\*}
 P(-T\_n\ge t)\le e^{-nI\_-(t)},
\end{equation\*}
where
\begin{equation\*}
 I\_-(t):=\sup\_{x\ge0}(tx-l(-x)), 
\end{equation\*}
so that
\begin{equation\*}
 P(|T\_n|\ge t)\le\min(1,e^{-nI\_+(t)}+e^{-nI\_-(t)}). \tag{3}
\end{equation\*}
Note that $I\_+(t)\ge th-l(h)\to\infty$ as $t\to\infty$, and similarly $I\_-(t)\to\infty$ as $t\to\infty$. Also, the functions $I\_\pm$ are nondecreasing and lower semi-continuous, being the pointwise suprema of a family of nondecreasing continuous functions; so, the functions $I\_\pm$ are also left-continuous. Also, $I\_\pm(0)=0$ -- because $l(0)=0$, $l'(0)=EY\_1=0$, and $l$ is convex, so that $l(x)\ge0$ for all $x\ge0$.
So, there exists
\begin{equation\*}
 t\_\*:=\max\{t\ge0\colon e^{-I\_+(t)}+e^{-I\_-(t)}\ge1\}\in(0,\infty), 
\end{equation\*}
and then $e^{-I\_+(t)}+e^{-I\_-(t)}\ge1$ for $t\in[0,t\_\*]$ and $e^{-I\_+(t)}+e^{-I\_-(t)}<1$ for $t\in(t\_\*,\infty)$.
Defining now the function $I$ by the requirement that
\begin{equation\*}
 e^{-I(t)}=e^{-I\_+(t)}+e^{-I\_-(t)}
\end{equation\*}
for $t>t\_\*$, with $I(t):=0$ for $t\in[0,t\_\*]$, we get
\begin{equation\*}
 \min(1,e^{-nI\_+(t)}+e^{-nI\_-(t)})\le e^{-nI(t)}
\end{equation\*}
for all natural $n$ and all real $t\ge0$.


Thus, (3) yields (1).


