Let $(X,\mathfrak{M},\mu)$ be a measure space with $\mu(X)=1$ and let $f$ be an $L^{\infty}(\mu)$ function on $X$. Prove that $$\lim_{t\to\infty}\frac{1}{t}\log\left(\int_{X}e^{tf}~d\mu\right)=\operatorname{ess-sup}_{x}f(x),$$ where $\operatorname{ess-sup}_{x}f(x)=\inf\left\lbrace t~:~\mu(\{x\in X~:~f(x)>t\})=0\right\rbrace$.
Since $f$ is essentially bounded, I know that $||f||_{\infty}<\infty$, so I know that the above limit should be finite. Should I be using Jensen's Inequality somewhere since $\log$ is concave and we are given $\mu(X)=1$? Thanks for any help.
Here are some hints.
We notice that $f(y)\leqslant \operatorname{ess-sup}_xf(x)$ hence $e^{tf(y)}\leqslant e^{t\operatorname{ess-sup}_xf(x)}$. Integrating, taking the $\log$, dividing by $1/t$ and letting $t$ going to infinity shows that $$ \limsup_{t\to +\infty}\frac{1}{t}\log\left(\int_{X}e^{tf}~d\mu\right)\leqslant\operatorname{ess-sup}_xf(x). $$
For an inequality with $\liminf$, define for a positive integer $n$ the set $$A_n:=\left\{y\in X, f(y)\geqslant\operatorname{ess-sup}_xf(x)-1/n \right\}.$$ Then $A_n$ has a positive measure and we can bound $\int_{X}e^{tf}~d\mu$ from below by the integral on $A_n$.