Lemma 2.2.4 (Jensen's inequality)9931E90187F6

Proof. For $u<v$ in $I$ put $s(u,v)=(\varphi (v)-\varphi (u))/(v-u)$. If $u<v<w$ lie in $I$, then $v=\lambda u+(1-\lambda )w$ with $\lambda =(w-v)/(w-u)\in (0,1)$, and the convexity inequality $\varphi (v)\leq \lambda \varphi (u)+(1-\lambda )\varphi (w)$ is equivalent to each of the inequalities $s(u,v)\leq s(u,w)$ and $s(u,w)\leq s(v,w)$. Hence $s(u,v)$ is nondecreasing in each of its two arguments. Consequently, if $[x,y]\subset I$ and $x'<x$, $y'>y$ are points of $I$, then $s(x',x)\leq s(u,v)\leq s(y,y')$ whenever $x\leq u<v\leq y$, so $\varphi $ is Lipschitz on $[x,y]$. In particular, $\varphi $ is continuous, and $\varphi \circ f$ is measurable.

Let $a=\int f\, \mathrm dP$. If $I$ has a finite left endpoint $\alpha $, then $f-\alpha >0$ on $X$, so one of the sets $\{ f-\alpha >1/k\} $, which increase to $X$, has positive measure, and therefore $a-\alpha =\int (f-\alpha )\, \mathrm dP>0$. In the same way, $a$ is smaller than a finite right endpoint of $I$; thus $a\in I$. For $u<a<w$ in $I$ we have $s(u,a)\leq s(u,w)\leq s(a,w)$, so $c=\sup _{u<a}s(u,a)$ is finite and $c\leq s(a,w)$ for every $w>a$. Hence $\varphi (y)\geq \varphi (a)+c(y-a)$ for every $y\in I$: for $y<a$ this is the inequality $s(y,a)\leq c$, and for $y>a$ it is $s(a,y)\geq c$. It follows that $\varphi \circ f\geq \varphi (a)+c(f-a)$, where the right side is integrable. Therefore the negative part of $\varphi \circ f$ is integrable, and since $\int (f-a)\, \mathrm dP=0$, integration gives $\int \varphi \circ f\, \mathrm dP\geq \varphi (a)$. $\square$