Lebesgue Integral: Convexity

Question

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?

Answer 1

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!

Answer 2

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: $$\int_\Omega f\mathrm{d}\rho\in\overline{\langle f(\Omega)\rangle}$$