# 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?

• How do you interpret the symbol $\int_\Omega f\,d\rho$ as a Riemann integral? – Martin Argerami Nov 9 '14 at 21:12
• It is meant to be the Lebesgue integral. From the Riemann integral it seems evident that the statement holds (for the Lebesgue integral). – C-Star-W-Star Nov 9 '14 at 21:18
• But what is the Riemann integral with respect to a probability measure? – Martin Argerami Nov 9 '14 at 21:19
• The limit of a net of the convex hull. – C-Star-W-Star Nov 9 '14 at 21:20
• But how to apply this to show the corresponding statement on Lebesgue integral? I mean not every Lebesgue integrable is Riemann integrable... – C-Star-W-Star Nov 9 '14 at 21:23

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!

• Ui ok let me have a look. :) (And yes, the braces were meant to denote the convex hull.) – C-Star-W-Star Nov 9 '14 at 22:00
• What about cases like $(a,\infty)$? – C-Star-W-Star Nov 9 '14 at 22:13
• Good question. I corrected the argument (it is now simpler than before ;) ). – PhoemueX Nov 10 '14 at 8:31
• There's also a problem when considering complex valued functions resp. functions over complex Banach spaces as splitting functions into real and imaginary part produces queries with the convex hulls... I managed to find a proof now for the Lebesgue integral of complex functions by an approximation argument. Can you have a look on it? (Like a check for raw mistakes...) – C-Star-W-Star Nov 10 '14 at 14:21
• You can just interpret every complex Banach space as a real Banach space. Likewise, the complex numbers are a real Banach space. – PhoemueX Nov 10 '14 at 16:55

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

• So, you are using $s_{N,n} = \sum_k b_{N,n,k} \chi_{A_{N,n,k}}$ and you choose $f_N ( \omega \in A_{N,n,k})$ as an arbitrary element of $f_N (A_{N,n,k})$? Also, you probably want to show (or you are using implicitly) that you can take $b_{N,n,k} \in f_N (A_{N,n,k})$. The general idea you are applying should work. – PhoemueX Nov 11 '14 at 14:37
• Yep, I used such simple functions $s_{N,n;k}=\sum_kb\chi$ but I directly chose a point $\omega\in A_{N,n;k}$ to get $\sum_kf_N(\omega)\rho(\ldots)$ as convex combination. Ok good. Do you think I could/should simplify notation since it seems pretty messy? – C-Star-W-Star Nov 11 '14 at 16:30