Prove that every nonempty open convex subset of $L^0([0,1])$ is the whole space. Let $(M,\mathcal{A},\mu)$ be a measurable space such that $\mu(M)<\infty$. Denote by $\mathcal{L}^0(\mu)$ the vector space of all real valued measurable functions on $M$, and define
$$
L^0(\mu):=\mathcal{L}^0(\mu)/\sim,
$$
where the equivalence relation is given by equality almost everywhere. The metric is
$$
d(f,g):=\int_M \frac{|f-g|}{1+|f-g|}d\mu.
$$
Take $M=[0,1]$. Prove that every nonempty open convex subset of $L^0([0,1])$ is the whole space.
Here is my attempt:
Let $C$ be an open convex subset of $L^0([0,1])$. Take $f\in L^0([0,1])-C$. We want to find $g_1,g_2\in C$ and a $\lambda\in(0,1)$ such that
$$
f=\lambda g_1+(1-\lambda) g_2.
$$
A trying is to take $\lambda=d(f,g)/\mu([0,1])$ for some $g\in C$, but it seems not working.
Could anyone give some hints?
 A: It is necessary to suppose that $\mu$ is atomless (otherwise, $\delta_0$ provides a counterexample). Now WLOG we can suppose that $\mu$ is a probability measure.
Let $B_r=\{f\in L^0:\|f\|<r\}$. It suffices to prove that, for every $r>0$,  $\text{conv}(B_r)=L^0([0,1])$. To do so, let $f\in L^0$.  For $n$ big enough, let $[x_k,x_k+1]$ be a uniform partition of $[0,1]$ in $n$ intervals. Define $g_k=nf(x)\chi_{[x_k,x_{k+1}]}$. Notice that
$$\|g_k\|=\int_{x_k}^{x_{k+1}}\frac{nf(x)}{1+nf(x)}d\mu(x)\le \mu([x_k,x_{k+1}])$$
Taking $n$ big enough, $\|g_k\|< r$ (here we need $\mu([0,1])=1$ and that $\mu$ is atomless) and so $g_k\in B_r$. This implies that $f=\sum_{k=1}^n \frac{1}{n}g_k\in \text{conv}(B_r)$ and so $\text{conv}(B_r)=L^0([0,1],\mu)$.
As a side note: this implies that there is no non-trivial continuous linear functional on $L^0(\mu)$.
A: I suppose $L^{0}[0,1]$ refers to the case when $\mu$ is the Lebesgue measure on $[0,1]$. However the argument below holds when Lebesgue measure is replaced by any continuous measure. [By continuous I mean $\mu \{x\}=0$ for all $x$. Any non-atomic measure has this property].
Wihtout loss of generality we may assuem that $0 \in C$. In this case some $\epsilon$ ball around $0$ is contained in $C$. Let $f\in L^{0}([0,1])$. The function $x\rightarrow \int\limits_{0}^{x}\left\vert
f(y)\right\vert ^{p}dy$ is continuous. Hence there exists $a\in (0,1)$ such
that $\int\limits_{0}^{a}\left\vert f(y)\right\vert ^{p}dy=\frac{1}{2}%
\int\limits_{0}^{1}\left\vert f(y)\right\vert ^{p}dy$. Let $%
f_{1}=2fI_{(0,a)}$ and $f_{2}=2fI_{[a,1)}$. Then $\frac{f_{1}+f_{2}}{2}=f$.
Also $\int \left\vert f_{1}\right\vert
^{p}=2^{p}\int\limits_{0}^{a}\left\vert f(y)\right\vert
^{p}dy=2^{p-1}\int\limits_{0}^{1}\left\vert f(y)\right\vert ^{p}dy$.
Similarly, $\int \left\vert f_{2}\right\vert
^{p}=2^{p-1}\int\limits_{0}^{1}\left\vert f(y)\right\vert ^{p}dy$. We have
expressed $f$ as a convex combination of functions $f_{1}$ and $f_{2}$ such
that $d(f_{j},0)=2^{p-1}d(f,0)$. Repeating this we can express $f$ as a
convex combination of a finite number of elements whose distance from $0$ is
as small as we want (in particular less than $\epsilon$) . ( Note that $p-1<0$). Hence $f \in C$.
