= Solution
Let $m=\mathbb EX$. First, $m$ lies in the open <interval> $I$. For example, if $I$ has a finite lower endpoint $\ell$, then $X-\ell>0$ <almost surely>, and its <expected value> is strictly positive; hence $m>\ell$. The corresponding argument at a finite upper endpoint gives the other strict inequality. <Integrability> ensures $m$ is finite when either endpoint is infinite.
The tangent <affine function> at $m$ gives the <supporting inequality for a differentiable convex function>. Under the stipulated condition $f''>0$, contact with that tangent occurs only at $m$: define
$$
h(t)=f(t)-f(m)-f'(m)(t-m).
$$
Then $h(m)=0$, and $h(t)>0$ for every $t\ne m$. One direct proof uses $h'(t)=f'(t)-f'(m)$: $f'$ is strictly increasing, so $h$ decreases towards $m$ on the left and increases away from $m$ on the right. Therefore
$$
\mathbb E[f(X)]=f(m)+\mathbb E[h(X)]\geq f(m),
$$
with equality exactly when $h(X)=0$ <almost surely>, equivalently when $X=m$ <almost surely>. Conversely, an <almost surely> constant $X=m$ plainly gives equality. This proves the <strict Jensen equality criterion>.
No extra assumption that $f(X)$ is an <integrable random variable> is needed to interpret this argument. Its <negative part> is bounded by the absolute value of the <integrable random variable> $f(m)+f'(m)(X-m)$, since $h(X)\geq0$. Thus its <expected value> is defined in $(-\infty,+\infty]$; if it equals the finite number $f(m)$, then $\mathbb E[h(X)]=0$ and the required <integrability> follows automatically. \b[Equality holds if and only if $X=\mathbb EX$ almost surely.]
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