# Prove that the set of extreme points in $B$ is equal to an atom

How to prove this? Please help me. Thank you very much.

A measurable set $E$ in a measure space $(X, \mathcal{M}, \mu)$ is said to be an atom if $\mu (E) > 0$ and no proper measurable subset of $E$ has positive $\mu$ measure.

Let $(X, \mathcal{M}, \mu)$ be a $\sigma$- finite measure space. Prove that the set of extreme points in $B_{L^1(\mu)}=\{v \in L^1(\mu): \|v\| \le 1\}$ is equal to $\{\pm \mu(F)^{-1} \chi_F: F \mbox{ is an atom of }\mu\}$.

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Under your unusual definition of an atom, this statement is false. Consider $X=[0,1]$ and $\mathcal M$ the $\sigma$-algebra generated by $X$ and its countable subsets, and let $\mu(A)=1$ if $A$ is uncountable and $\mu(A)=0$ if $A$ is countable. This $\sigma$-finite measure space has no atoms (since you can always remove further points from any set of positive measure and still leave a set of positive measure), yet $\pm\chi_X$ are extreme points of $B_{L^1(\mu)}$ (in fact the only points of $B_{L^1(\mu)}$).

The statement is true, however, under the usual definition of an atom (see e.g. Wikipedia and PlanetMath), under which an atom is a set $E$ with positive measure that has no subsets $S$ with $\mu(E)\gt\mu(S)\gt0$, i.e. all subsets of $E$ have measure either $0$ or $\mu(E)$. In the above example, $X$ is an atom under this definition, so things work out.

To prove the statement under this definition, first let $f$ be an extreme point of $B_{L^1(\mu)}$, and let $S$ be its support. If $S$ is not an atom, we can write it as a union $S=S_1\cup S_2$ of sets of positive measure. The integral of a positive integrable function over a set of positive measure is positive. (To see this, consider the preimages of $(1,\infty)$ and $(1/(n+1),1/n]$, which cannot all have measure zero.) Thus we can write $f$ as $f=I_1f_1+I_2f_2$ with

$$I_k=\int_{S_k}|f|\,\mathrm d\mu$$

and

$$f_k=\frac{f\cdot\chi_{S_k}}{I_k}\in B_{L^1(\mu)}\;,$$

contradicting the fact that $f$ is an extreme point. Hence $S$ is an atom. (Note that this argument wouldn't work under your definition of an atom, since in that case one of $S_1$ and $S_2$ could have zero measure, and then one of $I_1$ and $I_2$ would be zero and this wouldn't be a proper convex combination.)

Since $f$ is measurable, it must be constant on $S$ except for a set of zero measure: The intersections of the preimages $f^{-1}([x,\infty))$ with $S$ must all have measure $0$ or $\mu(S)$, so there must be some value $x_0$ were the jump occurs, and then $\mu(f^{-1}(\{x_0\}))=\mu(S)$. Thus $f$ is equivalent in the $L^1$ sense to a constant function on $S$, and since $\lVert f\rVert=1$, the constant must be $\pm\mu(S)^{-1}$. Thus all extreme points of $B_{L^1(\mu)}$ are of the given form.

I won't spell the other direction out in detail so as to leave some of your homework for you – if $\pm\mu(F)^{-1}\chi_F$ with $F$ an atom is a convex combination, you can show that the integrals over the absolute values of the summands outside $F$ must vanish, so the support of the summands is $F$ plus perhaps a set of measure zero, and by the above argument the summands have to be constant on $F$.

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Can you please elaborate more in the paragraph before the end? I dont see why f must be constant over S. – Danny Apr 23 '12 at 4:07
@Danny: There was a slight inaccuracy in that paragraph, which I've now corrected. Since $f$ is measurable, the preimages of the measurable sets $[x,\infty)$ must be measurable, and so must their intersections with $S$. Since $S$ is an atom, the measure of the intersections can only be $0$ or $\mu(S)$. This measure is a non-increasing function of $x$. For any non-increasing function of $x$ with exactly two values, there must be a point $x_0$ at which the function jumps from one value to the other. – joriki Apr 26 '12 at 11:47
If $\mu(f^{-1}([x_0,\infty))\cap S)=\mu(S)$ and $\mu(f^{-1}([x,\infty))\cap S)=0$ for all $x\gt x_0$, it follows that $\mu(f^{-1}(\{x_0\})=\mu(S)$ and $\mu(f^{-1}(\mathbb R\setminus\{x_0\})\cap S)=0$. Thus $f$ differs from a function constant on $S$ only on a set of measure zero, and it is therefore equivalent in the $L^1$ sense to a function constant on $S$. – joriki Apr 26 '12 at 11:47
Thanks a lot for your nice explanations, but I have another question. If I want to read more about extreme points and convex geometry, do you know any nice book for beginners? Any book? – Danny Apr 26 '12 at 22:09

Sorry to be pedantic; in the case of complex scalars, the proof shows that $|f|=t_{0}$ a.e. on $S$ (where $t_{0}=\frac{1}{\mu(S)}.$) In this case (e.g., Conway's Functional Analysis (2nd ed), page 144, problem 2), cover the (complex) unit circle with disjoint intervals $\{U_{i}: 1\le i\le n\}$ of size $\frac{2\pi}{n},$ notice that only one of $\{f^{-1}(t_{0}U_{i})\}$ ($=\{f^{-1}(V_{i})\},$ where $V_{i}=\bigcup_{t>0} tU_{i}$) has positive $\mu-$measure (since $S$ is an atom of $\mu$) and let $n\to\infty.$ (For beginners, Victor Klee's award winning article `What is a convex set?'' (http://www.maa.org/programs/maa-awards/writing-awards/what-is-a-convex-set) might be a good read.)

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