extreme points of unit balls on $L_1(\mu)$ I wanna show that for a measure space $(X, \mu)$, the set of extreme points of the unit ball of $L_1(\mu)$ equals the set of characteristic functions of atoms of $\mu$ multiplied by a suitable constant(to guarantee that the norm is 1).
I tried to show followings:
1) $f\in L_1(\mu)$, and if there are two disjoints subsets $A, B$ of $X$ with $A\cup B=X$ and $\int_A |f|d\mu>0, \int_B |f|d\mu>0$, then $f$ is not an extreme point of $L_1(\mu)$.
2)If there are no such pairs of $A, B$, then $f$ is a constant multiple of a characteristic function of an atom.
But it is hard to show 2). Is there any hints for this approach or other nice way to prove the original question?
 A: $(\mathcal{T}1)$ Every measurable function is constant on atoms.
Proof:
consider atom $A$ and function $f$ that isn't constant on $A$.There is a value $r\in \mathbb{R}$, such that $\mu(f^{-1}((-\infty,r)))>0\land \mu(f^{-1}([r,+\infty)))>0$. Those two sets are division of an atom and both have positive measure - contradiction. QED.
Conclusion: define $X_f:=\{x\in X:f(x)\neq 0\}$. If $\int_{X_f}fd\mu=1$, then every atom $A\subset X_f$ has finite measure. From now on I will consider only those.
By definition $X_f$ is an atom iff for any $A\subset X_f$ $A\cong\emptyset\lor A\cong X_f$, but it is also equivalent to statement: $$(*)\forall A\subset X_f:\int_A|f|d\mu=\int_{\emptyset}|f|d\mu\lor \int_A|f|d\mu=\int_{X_f}|f|d\mu$$
As you have noticed, all extreme points (if any) should be located on sphere, and existence of division $A,B$ such that $\int_A |f|d\mu>0, \int_B |f|d\mu>0$ is sufficient for $f$ to not be an extreme point (take $g_1:=\frac{1}{\mu(A)}\chi_Af$ and $g_2:=\frac{1}{\mu(B)}\chi_Bf$ then $f=\mu(A)g_1+\mu(B)g_2$, and $g_1\in \mathbb{S}, g_2\in \mathbb{S}$). In the light of (*) it could be stated: if $X_f$ is not atomic then $f$ is not an extreme point. Lets show that it is necessary too:
$(\mathcal{T}2)$ If $X_f$ is atomic, then $f$ is an extreme point.
Proof:
Suppose, that $f$ is not an extreme point: $$\exists{a\in (0,1)}\ \exists{g,h\in \mathbb{B}\setminus \{f\}}: f\cong ag+(1-a)h$$
It is clear, that $X_f\setminus (X_g \cup X_h)\cong \emptyset$ and in restriction to $X\setminus X_f$, $ag$ and $(1-a)h$ will cancel out:
$$ (ag+(1-a)h)\restriction_{X\setminus X_f}\cong 0=f\restriction_{X\setminus X_f}$$
but then
$$ (**)ag+(1-a)h\cong \chi_{X_f}(ag+(1-a)h)=a\chi_{X_f}g+(1-a)\chi_{X_f}h$$
 Because of $(\mathcal{T}1)$ functions $f,g$, and $h$ have to be (a.e.) constant within $X_f$:
$$(***)
\left\{
     \begin{array}{lr}
       g\restriction_{X_f}\cong uf\restriction_{X_f} \\
       h\restriction_{X_f}\cong vf\restriction_{X_f}
     \end{array}
   \right.
$$
for some $u,v\in [-1,1)$ ($u,v\neq 1$ because $g,h \neq f$). From $(**)$ and $(***)$:
$$ag+(1-a)h\cong a\chi_{X_f}uf+(1-a)\chi_{X_f}vf$$
$$ \chi_{X_f}f=f$$
$$ag+(1-a)h\cong auf+(1-a)vf\cong f$$
That boils down to one variable case: $1=au+(1-a)v$. There are no $u,v\in [-1,1)$ that satisfy this equation - contradiction.
Conclusion: $f$ is an extreme point. QED.
Notation: 


*

*$\chi_{\square}$ is characteristic function of the set $\square$

*$\mathbb{B}=\{f:X\to \mathbb{R}| f-measurable \land \int_{X}|f|d\mu\le 1\}$ is closed unit ball centered in constant zero function.

*$\mathbb{S}=\{f\in\mathbb{B}|\int_{X}|f|d\mu=1\}$ is unit sphere.

*$A\cong B$, $f\cong g$ means that $\mu(A\setminus B)+\mu(B\setminus A)=0$ and $\mu(\{x:f(x)\neq g(x)\})=0$ respectively.
