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For a continuous $F$ and a family satisfying a large deviation principle at speed $L$, the displayed exponential tail condition extends Varadhan's lemma to $F$ unbounded above. Apply the bounded-above result to $\min (F,M)$ and split the full integral into $F\le M$ and its tail; the tail is negligible and the truncated variational suprema increase to ${\sup }_x(F(x)-I(x))$. A sufficient condition is ${\mathrm{limsup}}_L{L}^{-1}\log \mathbb{E}{e}^{\gamma LF(X_L)}<\infty$ for some $\gamma >1$, since the tail is at most $e^{-(\gamma -1)LM}\mathbb{E}{e}^{\gamma LF(X_L)}$.

Let $Z_n$ clip the minimum of $k$ independent exponential sample means into $[a,b]$, with $0<a<1/\lambda <b$. The rate function under scalar clipping is the minimum cost on this interval. Apply Varadhan's lemma to $\log z$. On the lower side its objective is increasing; on the upper side its derivative is $(k+1)/z-k\lambda$. Thus the optimum is $z_{\ast }=(k+1)/(k\lambda )$ once $k\ge (\lambda b-1{)}^{-1}$, giving the displayed limit. The expectation is of the high power, rather than a power of the mean; the latter has limit $-\log \lambda$.

Varadhan's lemma states that if $U_n$ has a large deviation principle with good rate function $K$ and large-deviation speed $a_n$, then for every bounded continuous real function $h$,For a continuous function unbounded above, the same conclusion holds under the exponential tail extension of Varadhan's lemma, for example when the weighted contribution from $h>M$ is superexponentially negligible as $M\to \infty$.

The scalar clipping map $g(m)=\min (b,\max (a,m))$ is continuous. The rate function under scalar clipping has endpoint costs ${\inf }_{m\le a}J(m)$ and ${\inf }_{m\ge b}J(m)$. Since $a<\mu <b$ and the costs decrease towards $\mu$ from the left and increase to its right, both infima are attained at the corresponding endpoints. Thus the good rate function of $Z_n$ equals $J(z)$ on $[a,b]$ and is infinite outside. The function $\log z$ is bounded and continuous on that compact interval, soOn $[a,\mu ]$, the derivative of $\log z-I(z)$ is $2/z-\lambda >0$, so the maximum there is $\log \mu$. On $[\mu ,b]$, the derivative is $(k+1)/z-k\lambda$, with negative second derivative. Its unconstrained maximizer isIt lies in $[\mu ,b]$ whenever $k\ge (\lambda b-1{)}^{-1}$. The right-hand objective increases initially at $\mu$, so this maximizer dominates the whole left-hand interval. Substitution gives the limit for high moments of a clipped minimum of exponential sample meansThe moment here is the expectation of the $n$th power. If the printed notation were instead read as $(\mathbb{E}{Z}_n{)}^n$, its logarithmic limit would be $\log \mu =-\log \lambda$, since boundedness and convergence in probability give $\mathbb{E}{Z}_n\to \mu$.

A standard form of Varadhan's integral theorem is as follows. Let ${\mu }_L$ be probability measures on a Polish space $E$ satisfying a large deviation principle with speed $L\to \infty$ and good rate function $I$. If $F:E\to \mathbb{R}$ is continuous and bounded above, thenThe usual bounded-continuous version follows immediately. If $F$ is unbounded above, the same conclusion holds under the additional exponential tail conditionThis is the exponential tail extension of Varadhan's lemma. A convenient sufficient hypothesis isAn integrability hypothesis is essential for general unbounded functions; the ordinary large deviation principle alone does not control arbitrarily large exponential weights.

For example, let $X^L=L$ with probability $e^{-L^2}$ and $X^L=0$ otherwise. It satisfies the large deviation principle at speed $L$ with rate zero at zero and infinite elsewhere. But for $F(x)=x^2$, $\mathbb{E}{e}^{LF(X^L)}\ge e^{L^3-L^2}$, whose logarithmic rate diverges, while $\sup (F-I)=0$. This shows why the tail hypothesis cannot simply be omitted.

For the lower bound, fix $x$ with $I(x)<\infty$ and $\epsilon >0$. By continuity choose an open set $G$ containing $x$ on which $F(y)\ge F(x)-\epsilon$. ThenApply the open-set lower bound of the large deviation principle:Let $\epsilon \downarrow 0$ and take the supremum over $x$ to obtain the required lower bound. This argument also applies when $F$ is not bounded above.

For the upper bound, suppose $F\le M$ and set $S={\sup }_E(F-I)$, which is finite. Choose a large $K$ and partition $[-K,M]$ into finitely many closed intervals $[a_j,a_j+\epsilon ]$ of length at most $\epsilon$. The preimagesare closed sets. The overlapping endpoints do not matter for an upper estimate. Split off the lower tail and bound each remaining exponential weight:For a finite sum of nonnegative terms the upper logarithmic rate is at most the largest of their upper logarithmic rates. Applying the closed-set upper bound givesFor $x\in C_j$, $a_j+\epsilon \le F(x)+\epsilon$, hence every inner term is at most $S+\epsilon$. Empty preimages contribute no term. Let $K\to \infty$ and then $\epsilon \downarrow 0$ to get the upper bound $S$. Together with the lower bound this proves the bounded-above theorem.

For the unbounded extension, put $F_M=\min (F,M)$. This is continuous and bounded above, so the proved result applies, withThe full integral is at most the integral with $F_M$ plus the integral on $\{F>M\}$. The upper logarithmic rate is therefore at most the larger of $S_M$ and the tail rate. The tail hypothesis and then $M\to \infty$ give the upper bound $S$; the earlier local argument gives the lower bound. If $S=+\infty$, that lower argument already yields the asserted infinite limit.

Finally, the sufficient exponential-moment condition really implies the required tail estimate: on $F>M$,After integration the upper logarithmic rate is bounded by a finite constant minus $(\gamma -1)M$, which tends to $-\infty$. This completes the statement and proof of Varadhan's lemma, including the hypotheses needed for the unbounded version.