Show local convexity Given a metric space $(X,d)$ where $X=C([0,1])$ and $d$ is defined as: $$d(f,g)=\inf \{\epsilon : \mu \{x \in [0,1] : |f(x)-g(x)|>\epsilon\}<\epsilon\}$$ ($\mu$ is a Lebesgue measure).
I want to show this space is not locally convex. 
If it is not locally convex then $\exists u=\{g|d(f,g)<\epsilon \}$ in which $g_1, g_2\in u$ such that $d(f,rg_1+(1-r)g_2)>\epsilon,\ r\in [0,1]$. 
any hints on how to proceed?

It is easier to consider the space $Y$ of step functions, i.e., linear combinations of indicator functions of intervals with the same metric (which btw describes the convergence in measure). I claim that the only convex neighbourhood $U$ of $0$ is the full space $Y$: Indeed $U$ contains some ball $B_\varepsilon=\{f\in Y: \mu(\{t\in[0,1]: |f(t)|>\varepsilon\})<\varepsilon\}$. Given any $g\in Y$ you can write
$$
g=\frac{1}{n} \left( ngI_{[0,1/n]} + \sum_{k=1}^{n-1} ngI_{(k/n,(k+1)/n]}\right).
$$
If $n$ is bigger than $1/\varepsilon$, each term is in $B_\varepsilon \subseteq U$, and since $U$ is convex you get $g\in U$.
If you insist to have the space of continuous functions you can do the same with a continuous partition of unity. 
