Let $\mu$ be a probability measure on a measurable space $X$. Suppose that $\phi\colon X\to\mathbb{C}$ is integrable with respect to $\mu$. How does one prove that $$\int_{X}\phi \ \mathrm{d}\mu\in\overline{\mathrm{conv}(\phi(X))}?$$
So if $\phi$ is both non-negative and simple, then the integral is of the form $$\int_{X}\phi \ \mathrm{d}\mu=\sum_{j=1}^{n}\alpha_{j}\mu(B_{j})$$ where $\alpha_{1},\ldots,\alpha_{n}\in \phi(X)$ and $\sum_{j=1}^{n}\mu(B_{j})=1$. Hence in this case it is clear that $\int_{X}\phi \ \mathrm{d}\mu$ belongs to the convex hull of $\phi(X)$.
Since any non-negative function can be uniformly approximated by non-negative simple functions, it should follow that the integral of such a function belongs to the closure of the convex hull of $\phi(X)$.
In general we can write $\phi=\phi_{1}-\phi_{2}+i(\phi_{3}-\phi_{4})$ for non-negative $\phi_{1},\ldots,\phi_{4}$. But I don't see how the result follows from this.
Any suggestions would be greatly appreciated!
Let $A=\overline{ \text{conv}\phi(X)}$, $x_0= \int_X \phi\mathrm{d}\mu$. Suppose that $x_0 \notin A$, then there is a $\mathbb{R}$-linear functional $F: \mathbb{C} \rightarrow \mathbb{R}$ and a real number $c$ such that $F(x_0)<c$ and that $$F(x)>c \quad \forall x \in A.$$ Thus, $$c> F(x_0)= F\left( \int_X \phi\mathrm{d}\mu\right)= \int_X F\circ \phi \mathrm{d} \mu >c.$$ This is a contradiction. Hence, the conclusion.