Linear functionals on the space of measurable functions. Let us consider the following space of all measurable functions on $[0,1]$ with the metric: $$d(f, g) = \int_{0}^{1} \frac{|f-g|}{1 + |f-g|}d\mu.$$ It is the metric of convergence in measure on $[0,1]$ ($\mu$ - standard Lebesgue measure on $[0,1]$). I have problems with the proof that every continuous functional on this space is zero. I.e. there are no nontrivial continuous linear functionals on this (metric) space.
 A: We use the following

Proposition: Let $E$ be a topological vector space (not necessarily Hausdorff) over $\mathbb{R}$ or $\mathbb{C}$. There exists a nonzero continuous linear functional on $E$ if and only if there is a convex neighbourhood of $0$ that isn't the whole space.

Proof: Let $\lambda \neq 0$ be a continuous linear functional on $E$. Then $U = \lambda^{-1}(\{ z : \lvert z\rvert < 1\})$ is a convex and balanced proper subset of $E$. Since $\{ z : \lvert z\rvert < 1\}$ is open (in $\mathbb{R}$ or $\mathbb{C}$, as the case may be), and $\lambda$ is continuous, it follows that $U$ is open, and since $0\in U$, we see that $0$ has a nontrivial [meaning not the entire space] convex neighbourhood of $0$.
Conversely, suppose $V\neq E$ is a convex neighbourhood of $0$ in $E$. The Minowski functional
$$\mu_V \colon x \mapsto \inf \{ t > 0 : x \in tV\}$$
is then a sublinear functional (since $V$ is convex). Choose $x_0 \in E\setminus V$. By the Hahn-Banach extension lemma, we can find a real-linear functional $\lambda \colon E \to \mathbb{R}$ with $\lambda(x_0) = \mu_V(x_0)$ and $\lambda(x) \leqslant \mu_V(x)$ for all $x\in E$. Then $V \subset \lambda^{-1}\bigl((-1,1)\bigr)$, which shows that $\lambda$ is continuous. If $E$ is an $\mathbb{R}$-vector space, we're done, and if it's a complex vector space, let $\tilde{\lambda}(x) = \lambda(x) - i\lambda(ix)$ be the complex linear functional with real part $\lambda$. Clearly $\tilde{\lambda}$ is also continuous, and $\tilde{\lambda} \neq 0$.

Hence, to show that on a topological vector space no nonzero continuous linear functional exists, it suffices to show that the only convex neighbourhood of $0$ is the whole space. Equivalently, the convex hull of every neighbourhood of $0$ is the whole space.
Now, in the space $\mathscr{M}([0,1])$ of measurable functions on $[0,1]$, endowed with the pseudometric
$$d(f,g) = \int_0^1 \frac{\lvert f(x) - g(x)\rvert}{1 + \lvert f(x) - g(x)\rvert}\,dx$$
(we get a proper metric if we divide out the space of functions vanishing almost everywhere, but that's not necessary to do), let $V$ a neighbourhood of $0$. By definition of the topology, there is an $\varepsilon > 0$ such that
$$B_{\varepsilon}(0) = \{ f \in \mathscr{M}([0,1]) : d(f,0) < \varepsilon\} \subset V.$$
Choose $n \in \mathbb{N}$ such that $\frac{1}{n} < \varepsilon$. For $g \in \mathscr{M}([0,1])$ and $0 \leqslant k < n$, let
$$g_k(x) = \begin{cases} n\cdot g(x) &, \frac{k}{n} \leqslant x < \frac{k+1}{n} \\ \quad 0 &, \text{ otherwise} \end{cases}$$
(for $k = n-1$, let $g_{n-1}(x) = n\cdot g(x)$ on the closed interval $\bigl[\frac{n-1}{n},1\bigr]$ rather than on the half-open interval). Then we have
$$d(g_k,0) = \int_0^1 \frac{\lvert g_k(x)\rvert}{1 + \lvert g_k(x)\rvert}\,dx = \int_{k/n}^{(k+1)/n} \frac{n\lvert g(x)\rvert}{1 + n\lvert g(x)\rvert}\,dx < \int_{k/n}^{(k+1)/n} 1\,dx = \frac{1}{n} < \varepsilon,$$
and hence
$$g = \sum_{k = 0}^{n-1} \frac{1}{n}\cdot g_k$$
belongs to the convex hull of $B_{\varepsilon}(0)$.
Thus $\operatorname{conv} V = \mathscr{M}([0,1])$ for every neighbourhood of $0$, which by the proposition above implies that there is no nonzero continuous linear functional on $\mathscr{M}([0,1])$.
