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? 
 A: 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 $(f_{n_k})$.
A: 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 $A = \bigcup_{1<k\le N}A_k$ 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 $H \in L^1$.
