What do we know about the topological vector space of measurable functions?

Question

Suppose $(X,\mathcal{M},\mu)$ is a measure space, consider $$\mathcal{L}^0:=\{\text{all measurable functions}\ X\to\mathbb{C}\}/\sim$$where the equivalent relation is defined by $f\sim g $ iff $f=g\ a.e. $

Clearly $\mathcal{L}^0$ is a $\mathbb C$-vector space and there is a natural topology on it defined by convergence in measure, i.e., $$f_n\to f\ \ \text{iff}\ \ f_n\to f \ \ \text{in measure}$$

Under this convergence we can define cluster point hence closed and open set, and it's easy to prove that both addition and scalar multiplication are continuous under this topology, hence $\mathcal{L}^0$ is a topological vector space.

My question is : what do we know about this T.V.S. ? Is it locally convex ? Is it a Fréchet space ? This topology, defined by convergence, can it be induced by some metric,semimetric or seminorm? How to understand the term used in the definition of convergence in measure, i.e. $\mu\{x:|f_n-f|\ge \epsilon\}$, looks like to me it's a blueprint of some norm or metric...

Any hints or references are appreciated !

Edit 1): For the finite case $\mu(X)<\infty$, we already know this is a complete metric space with the metric defined by $$ d(f,g):=\int_{X} \frac{|f-g|} {1+|f-g| }d\mu $$ Also, in this case the convergence w.r.t. to the metric coincides with convergence in measure.

Note that this conclusion covers the case when $\mu$ is a probability measure, as discussed in Space of $L^0$ finite random variables. But for the general case it still puzzled me...

Edit 2): Recall that for measurable functions $f_n$ and $f$, $n=1,2,...$, we say $f_n\to f$ in measure provided for every $\epsilon>0$ , $\mu\{x:|f_n(x)-f(x)|\ge\epsilon\}\to 0$ as $n\to \infty$.

Answer 1

Let $(X,\mathcal{F},\mu)$ be a measure space and let $L_0$ denote the collection of all real-value $\mathcal{F}$-measurable functions $\mu$-mod $0$.

The posting here shows that there is a metric on $L_0$ that topologizes convergence in measure (in the broad sense defined in the OP). Two such metrics are described at the end of this posting.

Under such metrics $\rho$, it is easy to check that $(f,g)\mapsto f+g$ is continuous on $(L_0,\rho)\times (L_0,\rho)$ and thus, $L_0$ is a topological commutative group. Continuity of the scalar product $(\alpha,f)\mapsto \alpha f$ becomes an issue:

• If $\mu(X)<\infty$, the scalar product is continuous in which case, $L_0$ becomes a topological vector space.

• If $\mu(X)=\infty$, then continuity of the scalar product may fail. For example, consider $(\mathbb{R},\mathcal{M},\lambda)$, where $\mathcal{M}$ is the completion of the Borel $\sigma$-algebra under Lebesgue's measure $\lambda$. The sequence $f_n(x)=\frac1n x$ does not converge to $0$ in measure in the broad sense. More generally, if $\mu$ is an infinite semi finite space (i.e., $\mu(X)=\infty$ and for any $E\in\mathcal{F}$ with $\mu(E)=\infty$ there is $B\in \mathscr{M}$ with $0<\mu(B)<\infty$), one can show that the scalar product is not continuous.

• The issue is that the broad notion of convergence in measure in the OP is too coarse (or strong). For example, it is desirable that a sequence that converges a.s. also converges in measure. This does not happen in general when $\mu(X)=\infty$.

• Another notion of convergence in measure, more suitable for functional analysis, is the following: A sequence $f_n$ converges in measure to $f$ if for any set $A$ of $\mu$-finite measure, $$\lim_n\mu[A\cap(|f_n-f|>\varepsilon)]=0,\quad\forall\varepsilon>0$$ Notice that this version of convergence in measure coincided with the of the OP's when $\mu(X)<\infty$.

Under this finer notion, the family of metrics $(d_A:A\in L_1(\mu))$ on $L_0$ defined as $d_A(f,g)=\int_A|f-g|\wedge1\,d\mu$ make $L_0$ a topological vector space (see Fremlin, D. H., Measure Theory: broad foundations, vol 2., Torres Fremlin, 2001). When $\mu$ is $\sigma$-finite this finer version of convergence in measure admits a complete translation invariant metric under which $L_0$ becomes a topological vector space (but not necessarily a locally convex one).

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Metrization of course notion of convergence in measure

• Frechét metric: It is not difficult to show that $D:L_0\times L_0\mapsto\overline{\mathbb{R}}$ given by $$D(f,g)=\inf\{\mu(|f-g|\wedge1>\delta)+\delta:\delta>0\}$$ is in fact finite and that it defies a metric on $L_0$ where convergence in measure is equivalent to convergence in $D$ (even for infinite measure spaces).
  To see that $D(f,g)<\infty$ for any $f,g\in L_0$ notice that $1\in\{\mu\big(|f-g|\wedge1>\delta\big)+\delta:\delta>0\}$.
  To see that it topologies converges in measure in the broad sense defined by the OP:
  (Necessity) Suppose that $\lim_n\mu(|f_n-f|>\delta)=0$ for all $\delta>0$. Fix $\varepsilon>0$. There is $N$ such that for $n\geq N$ $$\mu(|f_n-f|\wedge1>\varepsilon)\leq \mu(|f_n-f|>\varepsilon)<\varepsilon$$ Hence, for $n\geq N$ $$D(f,f_n)\leq \mu(|f_n-f|\wedge1>\varepsilon)+\varepsilon<2\varepsilon$$ (Sufficiency) If $\lim_nD(f_n,f)=0$. Fix $\delta>0$ and choose $\varepsilon<\delta/2$. Then, there is $N$ such that $D_n(f,f_n)<\varepsilon$ whenever $n\geq N$. For each such $n$, choose $\varepsilon_n>0$ such that $$\mu(|f_n-f|\wedge1>\varepsilon_n)+\varepsilon_n<\varepsilon$$ It follows that $\varepsilon_n<\varepsilon$ and so, $$\mu(|f-f_n|>\delta)\leq \mu(|f-f_n|>\varepsilon_n)\leq \mu(|f-f_n|>\varepsilon_n)+\varepsilon_n<\varepsilon<\delta$$

• Ky Fan metric: The map $d:L_0\times L_0\rightarrow\mathbb{R}$ given by $$d(f,g)=\inf\{\delta>0: \mu(|f-g|\wedge1>\delta)\leq\delta\}$$ defines a metric on $L_0$ such that convergence in $d$ is equivalent to convergence in measure in the broad sense. A proof of the later fact, in the setting of finite measures, is given here. For general measure spaces, the proof goes verbatim once the usual distance $\rho(x,y)=|x-y|$ in $\mathbb{R}$ is substituted by the equivalent *but bounded distane $\rho'(x,y)=|x-y|\wedge1$.