Jensen's inequality for an integrable cost random variable

Put $m=\mathbb{E}U\ge a$. If $m=a$, let $V=(U-a{)}^{+}$, a nonnegative measurable variable equal to $U-a$ almost surely by [A1]. Thus $\mathbb{E}V=\mathbb{E}U-a=0$. Applying [F2] with the probability measure gives $V=0$ almost surely; combined with $U\ge a$ almost surely this implies $U=a$ almost surely. Consequently $\mathbb{E}f(U)=f(a)=f(m)$.

Suppose $m>a$. For $y>m$ define $r(y)=(f(y)-f(m))\mathrm{/}(y-m)$. Concavity applied to $a<m<y$ gives $r(y)\le (f(m)-f(a))\mathrm{/}(m-a)$, and the right-slope set is nonempty (take $y=m+1$) and bounded above; [F1] supplies its finite supremum $s$. For each $x$ with $a\le x<m$ and each $y>m$, concavity at $x<m<y$ gives $(f(m)-f(x))\mathrm{/}(m-x)\ge r(y)$; taking the supremum over $y$ yields $(f(m)-f(x))\mathrm{/}(m-x)\ge s$. Therefore every $u\ge a$ satisfies $f(u)\le f(m)+s(u-m)$.

The affine upper bound in step 1.2 is integrable because $U$ is integrable. Taking expectations gives $\mathbb{E}f(U)\le f(m)+s(\mathbb{E}U-m)=f(m)$. Together with step 1.1 this proves the claim.