Theorem: Let \(\varphi:\Bbb R \to \Bbb R\) be a convex function and \(f:[0,1]\to\Bbb R\) be an integrable function, and assume \(\varphi\circ f\) is integrable over [0,1]. Then

$$\varphi\left(\int_{[0,1]} f\right) \le \int_{[0,1]}\varphi \circ f $$

Proof: We prove this first for simple functions f, so let \(f = \sum_{k=1}^n c_k\chi_{E_k} \) for a finite collection of constants \(c_k\) and measurable sets \(E_k\), and in canonical form (so that in particular the sets form a partition of [0,1]).

Notice that in particular the assumptions above imply \(m([0,1]) = \sum_{k=1}^n m(E_k) = 1\). This is needed for us to apply the discrete version of Jensen's inequality,

$$ \begin{aligned}\varphi\left(\int_{[0,1]} f\right) &= \varphi\left(\sum_{k=1}^n c_km(E_k)\right) \\ &\le \sum_{k=1}^n m(E_k)\varphi(c_k) \\ &= \sum_{k=1}^n (\varphi\circ f)(c_k) m(E_k) \\ &= \int_{[0,1]} \varphi\circ f \end{aligned} $$

We use this result to prove the theorem for a general measurable f meeting the conditions of the theorem. We will crucially use the fact that every convex function on a bounded interval is absolutely continuous (and therefore continuous, and therefore the limit passes through \(\varphi\)). Moreover we also use the result that for every measurable function, there is a sequence of step functions \(\{s_n\}\) which approaches f from below. Therefore the Monotone Convergence Theorem ensures that the limit passes through the integral as well. Hence

$$ \begin{aligned}\varphi\left(\int_E f\right) &= \varphi\left(\int_E\lim_{n\to\infty}s_n\right) \\&= \lim_{n\to\infty}\varphi\left(\int_E s_n\right) \\&\le \lim_{n\to\infty}\int_E\varphi\circ s_n \\&= \int_E \varphi\circ (\lim_{n\to\infty}s_n) \\&= \int_E \varphi\circ f\end{aligned}$$