Topology of convergence in measure

Question

Currently I am doing some measure theory (on $X=[0,1]$ with the Borel-Sigma algebra and the Lebesgue measure), and I am looking at sets $A \subset L^p$, such that for all $q \in (0,p)$, the topologies induced by the $L^q$ and $L^p$ norm are equivalent. Okay, then my book said: Hey, if you know this for your set $A$, then you also know that the topologies induced by $L^p$ and $L^0$ agree on $A$ (and this is my main question: why?), where the latter topology is supposed to be induced by the convergence in measure. I am aware of the concept of convergence in measure from Stochastics, but I do not see a relationship to this problem.

Of course, I read the Wikipedia article and as far as I can see, this topology only defined for measure space that have a finite measure, right? Apparently, it is induced by a metric and so from a functional analytic point of view it would be interesting to know if this space is complete.

Nevertheless, I do not directly see the relationship between the definition of convergence in measure and what Wikipedia tries to define as the respective topology, is there anybody here who could elaborate on this? Also, I am puzzled in what sense we can think of this definition in terms of a formal limit of the $L^p$-norms. Is there a relationship that tells us, why this is the natural topology on $L^0$?

If anything is unclear, please let me know.

Answer 1

I read the relevant section of "A short course on Banach space theory", and I think you're misreading Carothers' claims. Carothers claims that if a (linear) subspace, not subset, $X$ of $L^{p}$, where $0<p\leq\infty$, is contained in a set $$M(p,\epsilon)=\left\{f\in L^{p}:m(\left|f\right|\geq\epsilon\left\|f\right\|_{p})\geq\epsilon\right\}$$ for some $0<\epsilon<1$, then for every $0<q<p$, the $L^{p}$ and $L^{q}$ norms are equivalent on $X$. This is Lemma 9.4. Topological equivalence on $X$ follows readily.

Going back to the subspace hypothesis, let $(f_{n})_{n=1}^{\infty}$ be a sequence in $X$ converging in $L^{q}$ norm to some $f\in L^{q}$. If $X$ is not a subspace, you don't necessarily know that $f-f_{n}\in X$ ($\forall n$) and therefore $$\left\|f-f_{n}\right\|_{q}\rightarrow 0\Rightarrow \left\|f-f_{n}\right\|_{p}\leq C_{q}\left\|f-f_{n}\right\|_{q}\rightarrow 0$$

To show that, under the same hypotheses, convergence in measure implies $L^{p}$ convergence, suppose that $\limsup \left\|f-f_{n}\right\|_{p}\geq\delta>0$. Passing to a subsequence, we may assume that $\left\|f-f_{n}\right\|_{p}\geq\delta$ for all $n$. Since $f-f_{n}\in M(p,\epsilon)$ for all $n$, we have that $$m\left(\left|f-f_{n}\right|\geq\delta\cdot\epsilon\right)\geq m\left(\left|f-f_{n}\right|\geq\left\|f-f_{n}\right\|_{p}\epsilon\right)\geq\epsilon \ (\forall n)$$ Hence, $f_{n}\not\rightarrow f$ in measure.

Answer 2

Suppose $(X,\mathcal M,\mu)$ is a finite measure space.

Let $\mathbb K\in \{\mathbb R,\mathbb C\}$ and let $M$ be the set of all measurable $\mathbb K$-valued functions on $X$ (w.r.t. our given measure and $\sigma$-algebra).

Given $(f_n)\in M^\mathbb N$ and $f\in M$, we say $(f_n)$ converges to $f$ in measure if for all $\epsilon>0$ we have $\mu(\{|f_n -f|\geq\epsilon\}\to 0$ as $n\to\infty$.

The function $d:M\times M\to \mathbb K$ defined by $d(f,g)=\int \frac {|f-g|}{1+|f-g|}d\mu $ is a metric. The assumption of finiteness of our measure space is there to guarantee this function is actually defined.

Theorem. $d$ makes $M$ into a complete metric space and convergence w.r.t. $d$ is equivalent to convergence in measure.

(are you asking for the proof of this theorem?)

Note: all of this may be generalized for $\sigma$-finite measures, if you make the definition of $d$ a bit more complicated.