# Completeness of a metric on measurable functions into a complete separable metric space

So this problem comes from Achim Klenke's Probability Theory: A comprehensive course that I am going through on my own. It appears as Exercise 6.2.1 in a chapter on convergence theorems. I have tried some angles but have not been able to crack it.

Let $$H \in \mathcal{L}^1(\mu)$$ with $$H > 0$$ $$\mu$$- almost everywhere and let $$(E,d)$$ be a separable complete metric space. For measurable $$f,g : \Omega \to E$$, define:

$$d_H(f,g) := \int_{\Omega} \min\{1 , d(f(\omega),g(\omega)) \} H(\omega) \ \mu(\mathrm{d}\omega)$$

i) Show that $$d_H$$ is a metric that induces convergence in measure

ii) Show that $$d_H$$ is complete if $$(E,d)$$ is complete.

I have been able to prove part (i) by splitting the integral up in various ways and using DCT and so on. But I have not been able to prove (ii). I thought the following Lemma proved earlier in the chapter could be useful,

Corollary 6.15 Let $$(E,d)$$ be a separable complete metric space. Let $$(f_n)_{n\in \mathbb{N}}$$ be a Cauchy sequence in measure in $$E$$, that is, for any $$A \in \mathcal{A}$$ with $$\mu(A) < \infty$$ and any $$\epsilon > 0$$ we have,

$$\mu(A \ \cap \ \{d(f_m, f_n) > \epsilon \}) \xrightarrow{\ m, n \to \infty } 0$$

Then $$(f_n)_{n\in \mathbb{N}}$$ converges in measure.

My idea was to prove that being Cauchy in $$d_H$$ entails being Cauchy in measure, which would then allow me have that $$(f_n)_{n\in\mathbb{N}}$$ converges in measure and so by part (i) it should converge in $$d_H$$ as well and hence we have that $$d_H$$ is complete.

One way I thought I could try to achieve this is by supposing we have some Cauchy sequence in $$d_H$$, that is, taking $$\epsilon > 0$$ arbitrary we have some $$N$$ such that,

$$d_H(f_m, f_n) < \epsilon, \ \forall \ m, n \geq N$$

Taking $$\delta > 0$$ arbitrary we seek to show that (assuming WLOG that $$\mu(\Omega) < \infty$$),

$$\mu(\{d(f_m, f_n) > \delta \}) < \epsilon$$

Or at least, some one-to-one function of $$\epsilon$$ (so that we can then pick the appropriate one when invoking the assumption of $$d_H$$ being Cauchy). Taking for now $$\delta \geq 1$$ we can see that in fact,

$$d_H(f_m, f_n) = \int_{\{d(f_m, f_n) \geq 1 \}} \min \{1, d(f,g) \}H \ \mathrm{d}\mu + \int_{\{d(f_m, f_n) < 1 \}} \min \{1, d(f,g) \}H \ \mathrm{d}\mu \leq \epsilon \\ \\ \Rightarrow \ \ \int_{\{d(f_m, f_n) \geq 1 \}} H \ \mathrm{d}\mu + \int_{\{d(f_m, f_n) < 1 \}} H d(f,g) \ \mathrm{d}\mu \leq \epsilon \\ \\ \Rightarrow \ \ \int_{\{d(f_m, f_n) \geq 1 \}} H \ \mathrm{d}\mu \leq \epsilon$$

At this point I wanted to use what I know about $$H$$ to try to get this inequality into one that uses $$\mu(\{d(f_m, f_n) \geq 1 \})$$ and the fact that $$H$$ is in $$\mathcal{L}^1$$ but I cannot think of how to do it. Intuitively there should be something I can do manipulate to separate the measure of the set the integral is being done over from the function. For instance if $$H = c$$ was a constant then we simply need to divide by that constant to get an estimate for the measure and hence allow us to conclude Cauchyness in measure. But I cannot think of how to proceed here. Am I on the right track?

For convergence in measure, assume $$d_H(f_n,f)\to 0$$. For fixed $$\epsilon \le 1$$ let $$E_n=\{\omega : d(f(\omega),g(\omega))\ge \epsilon\}$$ so $$\epsilon \chi_{E_n}\le \min(1,d(f(\omega),g(\omega)))$$ and therefore $$\int_{E_n} H\text{d}\mu \le \frac 1\epsilon d_H(f_n,f)$$. As you mentioned, if $$H\ge k$$ on $$\Omega$$ we would be done, but this need not be the case, however, we can show a similar property that is sufficient for our purpose. To wit, define $$A_k=\{\omega : \frac 1k \le H(\omega) \le \frac 1{k-1}\}$$ so $$\bigcup_{k >1} A_k \subset \Omega$$ implies $$\sum_{k>1} \mu(A_k)$$ converges. Hence, for every $$\delta>0$$ there exists some $$N$$ s.t. $$\sum_{k>N} \mu( A_k)<\delta$$. Moreover, $$H\ge \frac 1N$$ on $$A = \bigcup_{1 and $$\mu(A^c)<\delta$$. Therefore, \begin{align}\frac 1\epsilon d_H(f_n,f)\ge\int_{E_n} H\text{d}\mu&=\int_{A\cap E_n} H+\int_{A^c\cap E_n} H \\ &\ge \frac 1N\mu(A\cap E_n) \end{align} It follows that $$\mu(E_n)=\mu(E_n\cap A)+\mu(E_n\cap A^c)\le \mu(E_n\cap A)+\mu(A^c)\le \frac N\epsilon d_H(f_n,f)+\delta$$. Now choose $$n$$ large enough so $$d_H(f_n,f) \le \frac \epsilon N\delta$$ to get $$\mu(E_n)\le 2\delta$$, which shows $$d_H$$ induces convergence in measure.
Now, if $$f_n$$ is Cauchy, it is Cauchy in measure, and you can essentially adapt this answer (although Cor. 6.13 in Klenke's book is also useful) to show $$f_n \to f$$ in measure because $$d$$ is complete. Now let $$E = \{ \omega : d(f_n(\omega),f(\omega))\ge \epsilon\}$$ and compute \begin{align}d_H(f_n,f) &\le \int_EH\text{d}\mu +\epsilon\int_{E^c} H\text{d}\mu \\ & \le\epsilon +\epsilon||H||_1 \end{align} Whenever $$\mu(E)$$ is small enough because $$H \in L^1$$.
• "Now, if $(f_n)$ is Cauchy, it is Cauchy in measure". This is the exact part where I am having trouble accepting. We did prove that convergence in $d_H$ induces convergence in measure, but why does this mean that if $(f_n)$ is Cauchy in $d_H$ then it is Cauchy in measure? Is this supposed to be obvious and I am missing it? Thanks. – symchdmath Dec 16 '18 at 7:30
• @symchdmath yes, notice that when we prove that $\mu(E_n)\le \frac N\epsilon d_H(f_n,f)+\delta$, we could easily replace $f$ with $f_m$. So long as $n,m$ are large enough, if $f_n$ is Cauchy, $d_H(f_n,f_m)$ can be made less than $\frac \epsilon N \delta$, and Cauchy in measure still follows. Perhaps some relabeling of $E_n$ to $E_{n,m}$ would make it clearer, but the key fact remains that $d_H(f_n,f_m)$ can be made really small. – Guacho Perez Dec 16 '18 at 7:47
In any metric space if a Cauchy sequence has a convergent subsequence then the whole sequence converges. First choose an increasing sequence of integers $$n_k$$ such that $$\mu (d_H(f_{n_k},f_{n_j})>\frac 1 k)< \frac 1 k$$ for all $$j \geq k$$ and apply the corollary to the subsequence $$(f_{n_k})$$.
• Do you mean that $\mu(\{d(f_{n_k}, f_{n_j} > 1/k\}) < 1/k$? How are you sure that you can find such a sequence like that given only the assumption that $(f_n)$ is Cauchy in $d_H$. Thanks. – symchdmath Dec 16 '18 at 5:20
• @symchdmath because you proved that convergence in $d_H$ induces convergence in measure. – Guacho Perez Dec 16 '18 at 5:23
• Sorry I am not sure I am following fully. Given a Cauchy sequence in $d_H$ we have not proven that it converges as that is the point of the question. So how can we use that convergence in $d_H$ induces convergence in measure? Thanks. – symchdmath Dec 16 '18 at 6:13