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This is an exercise from Jacod's Probability Essentials: enter image description here

Putting everything into integration one can get:

$ \int_{\Omega}(\alpha-a)1_{A}dP\geq \int_{\Omega}[h(X)-a]dP \tag{*} $ where $(\Omega,\mathcal{A},P)$ is the underlying probability space and $ A:=\{\omega\in\Omega\mid h(X(\omega))\geq a\}. $ If $h$ is bounded by $\alpha$, then things can be done by (*). Any idea how I can go on?


[Remark] I didn't notice that $[0,\alpha]$ is the range of $h$ and thus $h$ is bounded by $\alpha$.

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Prove the pointwise inequality $ h(X)-a\leqslant(\alpha-a)\cdot\mathbf 1_A$ and integrate it with respect to $\mathrm P$. The LHS yields $\mathrm E(h(X))-a$ and the RHS yields $(\alpha-a)\cdot\mathrm P(A)$. This is the desired inequality.

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We have $E[h(X)]-a=\int_{\Omega}(h(\omega)-a)dP(\omega)=\int_A(h(\omega)-a)dP(\omega)+\int_{A^c}(h(\omega)-a)dP(\omega).$ On $A$, we have $h(\omega)-a\leq \alpha-a$ and on $A^c$ we have $h(\omega)-\alpha\leq 0$ hence $E[h(X)]-a\leq \int_A(h(\omega)-a)dP(\omega)=(\alpha-a)P(A),$ what is wanted.