Proof of Jensen's Inequality

Write $m=\int_{X}f\,d\mu$. Throughout, constants are integrated through $\mathbf{1}_{X}$, the map $X\to\mathbb{R}$ with value $1$ everywhere (Indicator Function of a Subset (i)): it is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), it lies in $\mathcal{M}^{+}$ with $\int_{X}\mathbf{1}_{X}\,d\mu=\mu(X)=1$ by Basic Properties of the Integral of Nonnegative Measurable Maps (ii) and Measure, Measure Space, and Probability Measure (v), and so it lies in $\mathcal{L}^{1}(\mu)$ with integral $1$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi). Hence, by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), for all $h\in\mathcal{L}^{1}(\mu)$ and $a,b\in\mathbb{R}$ the function $ah+b\mathbf{1}_{X}$, $x\mapsto ah(x)+b$, lies in $\mathcal{L}^{1}(\mu)$ and

$$\int_{X}\bigl(ah+b\mathbf{1}_{X}\bigr)\,d\mu=a\int_{X}h\,d\mu+b .$$

Step 1: the mean lies in $I$. Suppose first that no $p\in I$ satisfies $p<m$. Since $f(x)\in I$, then $f(x)\ge m$ for every $x\in X$. By the opening display the function $g=f-m\mathbf{1}_{X}$ lies in $\mathcal{L}^{1}(\mu)$ with $\int_{X}g\,d\mu=m-m=0$; it is measurable with nonnegative values, so by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi) it lies in $\mathcal{M}^{+}$ and its integral as an element of $\mathcal{M}^{+}$ is $0$. By The Lebesgue Integral and Null Sets (iv), $\mu(\{x\in X:g(x)>0\})=0$. Since $\mu(X)=1$, this set is not $X$, so there is $x_{0}\in X$ with $g(x_{0})=0$, that is, $m=f(x_{0})\in I$. In the same way, applied to $m\mathbf{1}_{X}-f$, if no $q\in I$ satisfies $q>m$ then $m\in I$. In the remaining case there are $p,q\in I$ with $p<m<q$, and $m\in I$ by Intervals of the Real Line (i). This proves (i).

Step 2: integrating a supporting line. Since $m\in I$, Convex Functions on an Open Interval: Monotone Slopes, Supporting Lines, Continuity, and the Supremum of Rational Supporting Lines (ii) provides the real number $c=c_{m}$, and Convex Functions on an Open Interval: Monotone Slopes, Supporting Lines, Continuity, and the Supremum of Rational Supporting Lines (iii), applied at $m$ and $f(x)\in I$, gives

$$\varphi\bigl(f(x)\bigr)\ge\varphi(m)+c\,\bigl(f(x)-m\bigr)=c\,f(x)+\bigl(\varphi(m)-c\,m\bigr)\qquad(x\in X).$$

By the opening display the right-hand side, as a function of $x$, is $cf+(\varphi(m)-cm)\mathbf{1}_{X}\in\mathcal{L}^{1}(\mu)$, with integral $c\,m+\varphi(m)-c\,m=\varphi(m)$. Since $\varphi\circ f\in\mathcal{L}^{1}(\mu)$, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix) gives

$$\varphi(m)=\int_{X}\bigl(cf+(\varphi(m)-cm)\mathbf{1}_{X}\bigr)\,d\mu\le\int_{X}\varphi\circ f\,d\mu,$$

which is (ii).