Lebesgue Integral: Convexity Given a finite measure $\mu(\Omega)<\infty$.
Consider a complex function $f\in\mathcal{L}(\rho)$.
From the Riemann integral it is evident that:
$$\frac{1}{\mu(\Omega)}\int_\Omega f\mathrm{d}\mu\in\overline{\langle f(\Omega)\rangle}$$
But how to prove this for the Lebesgue integral?
 A: Let $K := \overline{\langle f(\Omega) \rangle}$. It is closed and convex.
Assume $c := \frac{1}{\mu(\Omega)}\int f\mathrm{d}\mu\notin K$. It is compact and convex.
Then by geometric Hahn-Banach there exists a bounded linear functional with:
$$\varphi:E\to\mathbb{R}:\quad \varphi(c)< a\leq\varphi(K)$$
But that implies:
$$a =\frac{1}{\mu(\Omega)}\int a \mathrm{d}\mu \leq \frac{1}{\mu(\Omega)}\int \varphi(f) \mathrm{d}\mu = \varphi \left(\frac{1}{\mu(\Omega)}\int f \mathrm{d}\mu\right) < a\mu(\Omega)$$
That is a contradiction!
A: Consider the restricted, bounded case, first:
$$\Omega_N:=\{|f|\leq N\}:\quad f_N:=f\restriction_{\Omega_N}$$
(The restriction is necessary for later reasons.)
There exists a sequence of simple functions converging uniformly:
$$s_{N,n;k}=\sum_kb_{N,n;k}\chi_{A_{N,n;k}}:\quad\|f_N-s_{N,n}\|_\infty\to0$$
For they're Lebesgue integral one has:
$$\|\sum_kf_N(\omega\in A_{N,n;k})\rho(A_{N,n;k})-\sum_kb_{N,n;k}\rho(A_{N,n;k})\|\\\leq\|f_N-s_{N,n}\|_\infty\sum_k\rho(A_{N,n;k})=\|f_N-s_{N,n}\|_\infty\cdot\rho(\Omega_N)$$
Thus they tend to the closure of the convex hull:
$$\int_{\Omega_N}f_N\mathrm{d}\rho=\lim_n\int_{\Omega_N}s_{N,n}\mathrm{d}\rho\in\rho(\Omega_N)\overline{\langle f_N(\Omega_N)\rangle}$$
But the convex hulls agree on the restriction:
$$\langle f_N(\Omega_N)\rangle=\langle f(\Omega_N)\rangle\subseteq\langle f(\Omega)\rangle$$
(Here, the restriction was absolutely important to identify convex hulls.)
Now, consider the unrestricted, unbounded case.
By dominated convergence one has:
$$\int_{\Omega_N}f_N\mathrm{d}\rho=\int_\Omega f_N\mathrm{d}\rho\to\int_\Omega f\mathrm{d}\rho$$
and by continuity of the measure:
$$\rho(\Omega_N)\to\rho(\Omega)=1$$
Concluding that the assertion holds:
$$\int_\Omega f\mathrm{d}\rho\in\overline{\langle f(\Omega)\rangle}$$
