Bounty Edit: In the following, all the questions will be highlighted by a bold number and a text written in italics.
I found the following statement in a book, and I am really struggling to see why it is true.
The set of all measurable real functions on $[0,1]$, metrized via the map $$(f,g)\mapsto \int_{0}^1 \min \{1,|f(t)−g(t)|\}dt,$$ is a Fréchet space on which there is no nonzero continuous linear functional. (Even the linear functional $f\mapsto f(0)$ is not continuous on this space).
Regarding the very last statement in braces, I should have a proof of it.
Attempted Proof:
Let $f=0$, i.e. for every $x \in [0,1], f(x) = 0$, and define $f_n$ as $$ f_n (x):= \begin{cases} 1, & \text{ if } x \in [0, \frac{1}{n}],\\ 0, & \text{ if } x \in (\frac{1}{n}, 1].\\ \end{cases}$$ To see that $f_n \to f$, notice that, for an arbitrary $n \in \mathbb{N}$, we have $$ \begin{align*} d(f_n , f) &= \int_{0}^1 | f_n (t) - f(t) | \ dt \\ &\leq \int_{0}^1 | f_n (t)| + |f(t) | \ dt \\ &= \int_{0}^1 | f_n (t)|\ dt + \int_{0}^1 |f(t) | \ dt \\ &= \int_{0}^1 | f_n (t)|\ dt \to 0 \end{align*} $$ However, $f_n(0) \nrightarrow f(0)$, because for every $n \in \mathbb{N}$, $f_n (0) =1$, while $f(0)=0$.
Concerning it, here there is my first question:
1) Is the way in which I wrote the proof correct?
Regarding the main statement, the things that have to be proved are two:
a) $\mathcal{M}$ (that denotes the set of all measurable functions on $[0,1]$) is a Frechét space (complete metric linear space);
b) there are no nonzero continuous linear functionals on $\mathcal{M}$.
Related to part (b) I found in Rudin's "Functional Analysis" (pp.36-37, section 1.47) a proof that the space $L^p$ with $0 < p <1$ behaves in the same way. Moreover, on the wikipedia page on locally convex TVS I found that there is still another space that behaves similarly, namely $L_0$, i.e. the set of all measurable functions with distance function $$d(f,g):= \int_{0}^1 \frac{|f(x) - g(x)|}{1+|f(x)-g(x)|}dx.$$
Thus, here there are my other questions.
2) Is it possible to prove part (a) in different ways, namely one less constructive, and another one more constructive, with an actual $f \in \mathcal{M}$ to which an arbitrary Cauchy sequence $(f_n) \subseteq \mathcal{M}$ converge? If yes, how?
3) Is it possible to prove part (b) by showing that $\mathcal{M}$ is homeomorphic to $L_0$, which is homeomorphic to $L^p$ with $0 < p <1$, and then exploiting Rudin's proof? Are all those spaces actually homeomorphic? If yes, how?
4) If they are not homeomorphic, and we cannot use Rudin's proof, essentially, how do we prove the result, namely part (a) and part (b)?
I am looking forward to any answer, because being self-thaught I am having some real problems concerning this result, and I have the feeling that grasping this proof (with the techniques involved) would help me a lot.
Thank you for your time.
First a note on terminology: The more widely used convention is that a Fréchet space is a locally convex complete metrisable vector space. Complete metrisable vector spaces that are not (necessarily) locally convex may then be called $F$-spaces.
Next we have a problem in the substance matter. On the space of all measurable functions,
$$\rho \colon (f,g) \mapsto \int_0^1 \min \{ 1 , \lvert f(t) - g(t)\rvert\}\,dt$$
is not a metric, only a pseudometric. If two measurable functions are equal almost everywhere, we have $\rho (f,g) = 0$. To get a metric, we have to divide out the space $\mathcal{N}$ of almost everywhere vanishing functions. But if we do that, then $[f] \mapsto f(0)$ is not a well-defined map on $\mathcal{M}/\mathcal{N}$.
The same applies to the space $L_0$. Since for any $a \geqslant 0$ we have the inequalities
$$\frac{1}{2}\min \{ 1, a\} \leqslant \frac{a}{1+a} \leqslant \min \{ 1, a\},$$
we see that the pseudometrics $\rho$ and $d$ on the space of all measurable functions are Lipschitz-equivalent, in particular they induce the same uniform structure and hence the same topology. Thus we can in the following ignore $L_0$, everything said about $\mathcal{M}$ resp. $\mathcal{M}/\mathcal{N}$ applies - with perhaps minor changes in the formulation of inequalities involving the pseudometrics and of proofs - also to $L_0$ resp. $L_0/\mathcal{N}$.
However, there is no big problem in endowing $\mathcal{M}$ with the topology induced by the pseudometric $\rho$, it just isn't a Hausdorff space then. If we do that, the linear functional $f \mapsto f(0)$ is well-defined and discontinuous, and the answer to your question 1) is "Yes, your proof is correct". Although in that setting we could just take $f = 0$ and $g = \chi_{\{0\}}$, then we have $\rho (f,g) = 0$ but $f(0) = 0 \neq 1 = g(0)$.
$(\mathcal{M},\rho)$ is a complete pseudometric topological vector space
First, we need to prove that $\rho$ is a pseudometric. It is clear that we have $\rho (f,f) = 0$ for all $f\in \mathcal{M}$, and $\rho (f,g) \geqslant 0$ for all $f,g \in \mathcal{M}$. The symmetry $\rho (f,g) = \rho(g,f)$ follows from $\lvert -x\rvert = \lvert x\rvert$ for all $x\in \mathbb{R}$ (or $\mathbb{C}$), and the triangle inequality from the pointwise triangle inequality
$$\min \{ 1, \lvert a-c\rvert\} \leqslant \min \{ 1, \lvert a-b\rvert\} + \min \{ 1, \lvert b-c\rvert\}.$$
Next we note that $\rho$ is translation-invariant, for all $f,g,h\in \mathcal{M}$ we have $\rho (f,g) = \rho (f+h,g+h)$, and from that we deduce the (uniform) continuity of addition:
\begin{align} \rho (f+g,f_0+g_0) &= \rho\bigl((f-f_0) + (g-g_0),0\bigr)\\ &\leqslant \rho\bigl((f-f_0) + (g - g_0), g-g_0\bigr) + \rho(g-g_0,0)\\ &= \rho(f-f_0,0) + \rho(g-g_0,0)\\ &= \rho(f,f_0) + \rho(g,g_0), \end{align}
so $B_{\varepsilon/2}(f_0) + B_{\varepsilon/2}(g_0) \subset B_{\varepsilon}(f_0 + g_0)$.
To establish the continuity of scalar multiplication at the point $(\lambda_0,f_0) \in \mathbb{R}\times \mathcal{M}$, we write
$$\lambda f - \lambda_0 f_0 = (\lambda - \lambda_0) f_0 + \lambda(f-f_0).$$
Given $\varepsilon > 0$, we want to find $\delta,\eta > 0$ such that $\rho(\lambda f,\lambda_0 f_0) < \varepsilon$ whenever $\lvert\lambda - \lambda_0\rvert < \delta$ and $\rho(f,f_0) < \eta$. If we require $\delta \leqslant 1$, then it follows that $\lvert\lambda\rvert < 1 + \lvert\lambda_0\rvert$, and hence
$$\rho(\lambda f, \lambda f_0) \leqslant \max\{ 1, \lvert\lambda\rvert\}\cdot \rho(f,f_0) \leqslant (1 + \lvert\lambda_0\rvert)\cdot \rho(f,f_0).$$
Thus if we choose $\eta = \frac{\varepsilon}{2(1+\lvert\lambda_0\rvert)}$, we have
$$\rho(\lambda f,\lambda_0 f_0) < \rho\bigl((\lambda-\lambda_0) f_0,0\bigr) + \frac{\varepsilon}{2},$$
if $\rho(f,f_0) < \eta$ and $\lvert\lambda - \lambda_0\rvert < 1$.
It remains to show that we can choose $\delta \in (0,1]$ so that $\lvert\lambda - \lambda_0\rvert < \delta$ implies $\rho\bigl((\lambda - \lambda_0)f_0,0\bigr) < \frac{\varepsilon}{2}$. If that were not possible, there would exist a sequence $(\mu_n)$ with $\mu_n \to 0$ and $\rho(\mu_n f_0,0) \geqslant \frac{\varepsilon}{2}$ for all $n$. But $g_n(t) = \min \{ 1, \lvert \mu_n f_0(t)\rvert\}$ is a sequence of measurable functions with $0 \leqslant g_n(t) \leqslant 1$ for all $t$ and $g_n(t) \to 0$ for all $t$. The dominated convergence theorem asserts that
$$\rho(\mu_n f_0,0) = \int_0^1 g_n(t)\,dt \to 0,$$
contradicting our assumption. Hence scalar multiplication is continuous at $(\lambda_0, f_0)$.
So the topology induced by $\rho$ makes $\mathcal{M}$ a topological vector space.
Next, let's show completeness. It suffices to show that every Cauchy sequence has a convergent subsequence, since Cauchy sequences converge to all of their cluster points (and since we have a single pseudometric, it suffices to consider Cauchy sequences rather than Cauchy filters/Cauchy nets). Before proving the completeness, we note that pointwise almost everywhere convergence of a sequence of measurable functions implies $\rho$-convergence by the dominated convergence theorem.
Now let $(f_n)$ be a $\rho$-Cauchy sequence in $\mathcal{M}$. By extracting a subsequence, we may assume that $\rho(f_n, f_{n+1}) < 4^{-n}$ for all $n$. Then we define $A_n = \{ t \in [0,1] : \lvert f_n(t) - f_{n+1}(t)\rvert \geqslant 2^{-n}\}$ and note
$$4^{-n} > \rho(f_n, f_{n+1}) \geqslant \int_{A_n} \min \{ 1, \lvert f_n(t) - f_{n+1}(t)\rvert\}\,dt \geqslant \int_{A_n} 2^{-n}\,dt = 2^{-n}\lambda(A_n),$$
whence $\lambda(A_n) < 2^{-n}$, where $\lambda$ denotes the Lebesgue measure. Since $\sum \lambda(A_n) < +\infty$, the Borel-Cantelli lemma asserts that $E = \limsup\limits_{n\to \infty} A_n = \{ t \in [0,1] : t\in A_n \text{ for infinitely many } n\}$ is a null set. For all $t\in [0,1]\setminus E$, the sequence $\bigl(f_n(t)\bigr)$ is a Cauchy sequence, hence
$$f(t) = \begin{cases} \quad 0 &, t \in E \\ \lim\limits_{n\to\infty} f_n(t) &, t \notin E \end{cases}$$
is a measurable function, and $f_n \to f$ pointwise almost everywhere, so $\rho(f_n,f) \to 0$, which shows the completeness.
Existence of continuous linear functionals
If $E$ is a topological vector space, and $\lambda\neq 0$ a continuous linear functional on $E$, then $E$ possesses nonempty open convex proper subsets - for every nonempty open convex proper subset $U$ of the scalar field (be that $\mathbb{R}$ or $\mathbb{C}$), $\lambda^{-1}(U)$ is a nonempty open convex proper subset. Conversely, the Hahn-Banach theorem asserts the existence of a nonzero continuous linear functional on $E$ if $E$ possesses a nonempty open convex proper subset.
Since translations are homeomorphisms in topological vector spaces and they preserve convexity, the nonexistence of nonempty open convex proper subsets is equivalent to the assertion that the convex hull of every neighbourhood of $0$ is the whole space. This is often the easiest way to prove that a TVS admits no nonzero continuous linear functional.
Let $f\in \mathcal{M}$ be arbitrary. For $n \in \mathbb{N}\setminus \{0\}$, we define $I^{(n)}_k = \bigl(\frac{k-1}{n},\frac{k}{n}\bigr]$ for $2 \leqslant k \leqslant n$ and $I^{(n)}_1 = \bigl[0,\frac{1}{n}\bigr]$. Also, we set $g^{(n)}_k = n\cdot f\cdot \chi_{I^{(n)}_k}$.
Then $\rho(g^{(n)}{k},0) \leqslant \frac{1}{n}$, and the identity $f = \sum_{k = 1}^n \frac{1}{n}\cdot g^{(n)}_k$ shows $f \in \operatorname{co} B_{\varepsilon}(0)$ when we pick $n > 1/\varepsilon$.
The argument for the $L^p$ spaces with $0 < p < 1$ is similar, but we need to choose the partitions of $[0,1]$ based on $f$. Since
$$x \mapsto \int_0^x \lvert f(t)\rvert^p\,dt$$
is a continuous nondecreasing function, for every $n\in \mathbb{N}\setminus \{0\}$ we find $0 = x_0 < x_1 < \dotsc < x_{n-1} < x_n = 1$ with
$$\int_0^{x_k} \lvert f(t)\rvert^p\,dt = \frac{k}{n}\int_0^1 \lvert f(t)\rvert^p\,dt.$$
Similar to the above, we define $I^{(n)}_k = (x_{k-1}, x_k]$ (resp. $I^{(n)}_1 = [0,x_1]$) and $g^{(n)}_k = n \cdot f \cdot \chi_{I^{(n)}_k}$. Then
$$\int_0^1 \lvert g^{(n)}_k(t)\rvert^p\,dt = n^{p-1}\int_0^1 \lvert f(t)\rvert^p\,dt,$$
and since $p-1 < 0$, we can choose $n$ so large that $f = \sum_{k = 1}^n \frac{1}{n} g^{(n)}_k$ represents $f$ as a convex combination of elements of $B_{\varepsilon}(0)$, whatever $\varepsilon > 0$ is given.
Nonexistence of isomorphisms between $\mathcal{M}/\mathcal{N}$ and $L^p$
I don't know whether any of these spaces are homeomorphic, but if we restrict our attention to linear homeomorphisms (aka isomaorphisms in the category of topological vector spaces [over $\mathbb{R}$ resp. $\mathbb{C}$]), which is probably intended, we can see that $\mathcal{M}/\mathcal{N}$ is not isomorphic to $L^p$ for any $p$ (nor is $\mathcal{M}$ isomorphic to $\mathcal{L}^p([0,1])$, the space of all $p$-integrable measurable functions).
Namely, all $L^p$ (resp. $\mathcal{L}^p$) spaces are locally bounded, while $\mathcal{M}/\mathcal{N}$ (resp. $\mathcal{M}$) is not locally bounded.
By the $p$-homogeneity of the distance on $L^p$ (resp. $\mathcal{L}^p$), we have $c\cdot B_r(0) = B_{c^p\cdot r}(0)$ for all $c > 0$, so all balls with finite radius in $L^p$ ($\mathcal{L}^p$) are bounded [in the sense of TVSs].
To see that $\mathcal{M}/\mathcal{N}$ ($\mathcal{M}$) is not locally bounded, let $0 < r_1 < s < r_2 \leqslant 1$. Then we have $\rho(\lambda\cdot \chi_{[0,s]},0) \leqslant s < r_2$ for every $\lambda \in \mathbb{R}$, so $B_{r_2}(0)$ contains the line $\mathbb{R}\cdot \chi_{[0,s]}$, hence every multiple $c\cdot B_{r_2}(0)$ with $c\neq 0$ also contains that whole line. But $\rho(\chi_{[0,s]},0) = s > r_1$, so $B_{r_1}(0)$ does not contain the whole line, hence no ball $B_r(0)$ is bounded in $\mathcal{M}/\mathcal{N}$ ($\mathcal{M}$).
Since linear homeomorphisms preserve local boundedness, it follows that $\mathcal{M}/\mathcal{N}$ is not (topologically) isomorphic to any $L^p$ space.
Answers to the further questions
1) was already mentioned, $(f,g) \mapsto \int_0^1 \min \{ 1, \lvert f(t) - g(t)\rvert\}\,dt$ is not actually a metric but only a pseudometric on $\mathcal{M}$, and on the associated metric TVS $\mathcal{M}/\mathcal{N}$, the map $[f] \mapsto f(0)$ is not well-defined. Your proof is correct as a proof that the linear functional $f \mapsto f(0)$ on $(\mathcal{M},\rho)$ is not continuous.
2) I'm not sure what exactly "Is it possible to prove part (a) in different ways" is asking.
If it's about the fact that $(\mathcal{M},\rho)$ is a complete pseudometric topological vector space, then of course there are possible variations in the proof, but I don't see essentially different ways to go about it. In particular, "one less constructive, and another one more constructive, with an actual $f \in \mathcal{M}$ to which an arbitrary Cauchy sequence $(f_n) \subseteq \mathcal{M}$ converges", for the completeness, one needs to consider an arbitrary Cauchy sequence, so one can't really have something concrete.
If it's about the nonexistence of nonzero continuous linear functionals, we again face the situation that we need to deal with all nonzero linear functionals and show that they aren't continuous. Since there are too many linear functionals, and they can't be easily described, we can't be concrete here either, although for every given nonzero linear functional it is probably not too difficult to explicitly show its discontinuity with a concrete example.
3) As mentioned, $\mathcal{M}$ is the same as $L_0$, the two pseudometrics are Lipschitz-equivalent, so showing that one doesn't admit continuous nonzero linear functionals shows that the other doesn't admit them either. Since $\mathcal{M}/\mathcal{N}$ is not isomorphic (as TVSs) to any $L^p$, we can't directly prove (b) using Rudin's proof of the corresponding feature of the $L^p$ spaces, but we can use the same idea. In fact, as it turns out the construction is slightly easier for $\mathcal{M}$.
4) See above. We cannot directly use Rudin's proof, but the proof of (b) is an easy modification. The proof of completeness is a little different than for the $L^p$ spaces, because due to the capping of the integrand at $1$, it is not as easy to see that an absolutely convergent series in $\mathcal{M}$ must be pointwise convergent almost everywhere, we invoked the Borel-Cantelli lemma for that.