Let $L_0([0,1]) = \{ f : [0,1] \to \mathbb{R} : f \text{ is measurable} \}$. We define $$ d(f, g) = \int_0^1 \frac{|f(x) - g(x)|}{1 + |f(x) - g(x)|} \, dx. $$
I am interested in proving that:
$d$ generates on $L_0([0,1])$ a topology of a topological vector space, where convergence is equivalent to convergence in measure.
the only non-empty open convex set is $L_0([0,1])$ so $L_0([0,1])$ with the topology induced by the metric is not a locally convex space.
My attempt
In order to see that $d$ generates on $L_0([0,1])$ a topology of a topological vector space, let us prove that addition and scalar multiplication are continuous.
Let $f,g \in L_0([0,1])$ and $V$ a neighborhood of $f+g$. Then there must exist neighborhoods $V_1$ and $V_2$ of $f$ and $g$ respectively such that $V_1 + V_2 \subset V$. As $L_0([0,1])$ is a metric space, this is equivalent to say that given $\varepsilon$ we can find $\delta$ such that if $h_1 \in B_d(f,\delta)$ and $h_2 \in B_d(g,\delta)$, then $h_1+h_2 \in B_d(f+g, \varepsilon)$. We have $$ d(f+g, h_1+h_2) = \int_0^1 \frac{|(f+g)(x) - (h_1+h_2)(x)|}{1 + |(f+g)(x) - (h_1+h_2)(x)|} \, dx.$$ Now, noting that the function $\phi(t) = \frac{t}{1+t}$ is non-decreasing in $[0,1]$ and $$ |(f+g)(x) - (h_1+h_2)(x)| \leq |f(x)-h_1(x)| + |g(x)-h_2(x)|, $$ if we take $\delta = \frac{\varepsilon}{2}$, we deduce \begin{equation} \nonumber \begin{split} & d(f+g, h_1+h_2) \leq \int_0^1 \frac{ |f(x)-h_1(x)| + |g(x)-h_2(x)|}{1 + |f(x)-h_1(x)| + |g(x)-h_2(x)|} \, dx \\ &\leq \int_0^1 \frac{ |f(x)-h_1(x)| }{1 + |f(x)-h_1(x)| + |g(x)-h_2(x)|} \, dx + \int_0^1 \frac{ |g(x)-h_2(x)| }{1 + |f(x)-h_1(x)| + |g(x)-h_2(x)|} \, dx \\ &\leq \int_0^1 \frac{ |f(x)-h_1(x)| }{1 + |f(x)-h_1(x)| } \, dx + \int_0^1 \frac{ |g(x)-h_2(x)| }{1 + |g(x)-h_2(x)|} \, dx \leq \varepsilon. \end{split} \end{equation}
For scalar multiplication, given $f \in V = L_0([0,1]), \lambda$ a scalar and $V$ a neighbourhood of $\lambda f$, we want to prove that there exists $\delta$ such that if $\alpha \in (\lambda - \delta, \lambda+\delta)$ and $g \in B_d(f, \delta)$ then $\alpha g \in B_d(\lambda f, \varepsilon).$ However, I am stuck at this step:
\begin{equation} \nonumber \begin{split} d(\alpha g, \lambda f) &= \int_0^1 \frac{ |\alpha g(x) - \lambda f(x)|}{1 + |\alpha g(x) - \lambda f(x)|} \, dx = \int_0^1 \frac{ |\alpha g(x) - \alpha f(x) + \alpha f(x) - \lambda f(x)|}{1 + |\alpha g(x) - \alpha f(x) + \alpha f(x) - \lambda f(x)|} \, dx \\ & \leq \int_0^1 \frac{ |\alpha| |g(x) - f(x)|}{ 1 + |\alpha| |g(x) - f(x)|} \, dx + \int_0^1 \frac{ |f(x)| |\alpha - \lambda|}{ 1 + |f(x)| |\alpha - \lambda|} \, dx. \end{split} \end{equation}
- Let $V$ be a convex neighborhood of zero, we want to prove that $V = L_0([0,1])$. We know that there exists $r>0$ such that $B_d(0,r) \subset V$. Now, take $f \in L_0([0,1])$, do we have that $f \in V$? My guess is to try to write $f$ as a convex combination of elements of $ B_d(0,r)$ in the following way: $$ f = \frac{g_1 + ... + g_n}{n}.$$ I would make a partition of the interval $\{ x_0 = 0,..., x_n = 1 \}$ and define $g_i = \chi_{[x_{i-1}, x_i]} n \, f$. As the function $$ F(x) = \int_0^x \frac{|f(x)|}{1 + |f(x) |} \, dx$$ is continuous and $F(0) = 0 $ and $F(1) = d(0,f) $, it takes all posible values between $0$ and $d(0,f)$ so we can define $$ \int_{x_{i-1}}^{x_{i}} \frac{|f(x)|}{1 + |f(x)|} \, dx = \frac{d(0,f)}{n}. $$ However, do we have that $$d(0,g_i) = \int_{x_{i-1}}^{x_{i}} \frac{n|f(x)|}{1 + n|f(x)|} \, dx < r?$$
