# A question about measure-weighted barycenters

This is a question taken (inferred) from Ex. 19, chap 3 in Rudin's Real and Complex Analysis book.

Let $$\mu$$ be Legesgue's measure on $$X=[0,1]$$. Given a measurable $$L^{\infty}$$ function $$f:X\to C$$, we denote by $$R_f$$ the essential image of $$f$$ (that is, the set of complex numbers $$z\in C$$ such that $$\mu(f^{-1}(D(z,\epsilon))>0$$ for all $$\epsilon>0$$, with $$D(z,\epsilon)$$ being the disc of center $$z$$ and radius $$\epsilon$$). We also denote by $$A_f$$ the set of all numbers of the form $$z_E=\mu(E)^{-1}\int_E f\,d\mu$$ for some measurable subset $$E\subset [0,1]$$ with $$\mu(E)>0$$.

It can be shown (not too difficult) that $$R_f$$ is a closed compact set and that we always have the inclusions $$R_f\subset \bar{A_f}=\operatorname{Conv}(R_f)$$, where $$\operatorname{Conv}(R_f)$$ denotes the convex envelope of $$R_f$$. (This property holds, I think, more generally for any finite, diffuse measure on a locally compact space.)

The question is: is it true that for any $$f\in L^{\infty}([0,1],\mu)$$, the set $$A_f$$ is actually convex, and if so, how to prove it?

• Sorry for my mistake, $E_f$ should have read $R_f$ above (I hope this is corrected) – jacaboul Jun 5 '19 at 16:47
• What have you tried? – Praphulla Koushik Jun 5 '19 at 16:49
• All the examples that I could think of, so far, gives a convex $A_f$. Note for example that if $f$ has a finite essential image, then the result is easy to prove. – jacaboul Jun 5 '19 at 18:47
• The problem is that in general there might be many $z\in R_f$ that are not of the form $z_E$ for some $E$, but intuitively those points should lie on the boundary of ${\rm Conv} R_f$. More precisely I suspect one could show that $A_f$ must be locally closed (from which the result would follow), but I am so far unable to prove it. – jacaboul Jun 5 '19 at 18:54

Claim: For every $$f \in L^1[0,1]$$, the set $$A_f$$ is convex.

Proof: Let $$\mu_1=\mu$$ be Lebesgue measure on $$[0,1]$$ and consider the signed measures $$\mu_2,\mu_3$$ on $$[0,1]$$ defined using the real and imaginary parts of $$f$$ by $$\mu_2(E)= \int_E {\mathrm Re} (f) \,d\mu$$ and $$\mu_3(E)= \int_E {\mathrm Im} (f) \,d\mu$$.

We are given Lebesgue measurable sets $$D,E$$ in $$[0,1]$$ and $$a,b>0$$ with $$a+b=1$$. We must show there is a measurable set $$G$$ with $$(*) \quad z_G=a \cdot z_D+b \cdot z_E \, .$$ We may assume that $$r:=\mu(D)/\mu(E) \le 1$$. By the Lyapunov theorem on convexity of the range of nonatomic vector measures (see references below; it applies to signed measures, see e.g. ) there is a measurable set $$E_1$$ in $$[0,1]$$ with $$(\mu_1,\mu_2,\mu_3)(E_1)=r \cdot (\mu_1,\mu_2,\mu_3)(E) +(1-r) \cdot(\mu_1,\mu_2,\mu_3)(\emptyset) \, ,$$ so $$\mu(E_1)=\mu(D)$$ and $$z_{E_1}=z_E$$. By another application of Lyapunov's theorem, there is a measurable set $$G$$ in $$[0,1]$$ with $$(\mu_1,\mu_2,\mu_3)(G)=a \cdot(\mu_1,\mu_2,\mu_3)(D)+b \cdot (\mu_1,\mu_2,\mu_3)(E_1) \,.$$ Clearly $$\mu(G)=\mu(D)=\mu(E_1)$$ and $$G$$ satisfies (*).

References:

• Many thanks for this elegant proof, and for pointing out the link to Lyapunov's theorem. There is one misprint above: the factors $\mu(E)^{-1}$ in the definition of $\mu_2$ and $\mu_3$ are to be removed. – jacaboul Jun 6 '19 at 16:40