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Let $(X,\mathcal{M},\mu)$ be a $\sigma$-finite measure space and $f$ a measurable real valued function on $X$. Prove that \begin{equation*} \int_X e^{f(x)}\mathrm{d}\mu(x) = \int_\mathbb{R} e^{t}\mu(E_t)\mathrm{d}t \end{equation*} where $E_t=\{x\mid f(x)>t\}$ for each $t\in\mathbb{R}$.

Can this be solved by a change of variable formula?

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    Probably you should prove it when $f$ is simple first, then take limits – 2017-01-02
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    Write $\mu(E_t) $ as the integral of a characteristic/indicator function and use Fubini. – 2017-01-02
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    @user6246 Thank you for the suggestion. I think we can try $f=\sum_{n=1}^N a_n\chi_{E_n}$, where the sets $E_n$'s are disjoint. It follows that $e^{f(t)}$ is also a simple function, $e^{f(t)}=\sum_{n=1}^N e^{a_n}\chi_{E_n}$. Then $\int_X e^{f(x)}\mathrm{d}\mu(x)=\sum_{n=1}^N e^{a_n}\mu(E_n)$. In this case, $E_t=\{x\mid f(x)>t\}=\bigcup\limits_{\substack{n\\a_n>t}} E_n$. Hence $\mu(E_t)=\sum\limits_{\substack{n\\a_n>t}}\mu(E_n)$. I don't think we have $\int_X e^{f(x)}\mathrm{d}\mu(x)=\sum_{n=1}^N e^{a_n}\mu(E_n)=\int_\mathbb{R}e^t \mu(E_t)\mathrm{d}t$ yet. – 2017-01-02
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    @khalatnikov Fubini's theorem is a much better idea. You can do it for simple functions directly but using the same labels $E_n$ and $E_t$ for the two sets is not helpful. – 2017-01-02
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    @user6246 Thank you. You are quite right. The symbols should have been more distinct. – 2017-01-02

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\begin{align*} \int_{X}e^{f(x)}d\mu(x)& = \int_{X}\int_{-\infty}^{f(x)}e^{t}dt d\mu(x)\\ & = \int_{X}\int_{\mathbb{R}}I_{\{t< f(x)\}}(t)e^{t}dtd\mu(x)\\ & = \int_{X}\int_{\mathbb{R}}I_{\{f(x)>t\}}(x)e^{t}dtd\mu(x)\\ & = \int_{\mathbb{R}}\int_{X}I_{\{f(x)>t\}}(x)d\mu(x)e^tdt\\ & = \int_{\mathbb{R}}e^t\mu\left\{ f(x)>t \right\}dt. \end{align*}

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    The fourth equality is because of Fubini's theorem. – 2017-01-02