Convergence in measure iff convergence in distance metric

I am trying to solve the following question:

Let $(X, F, \mu)$ be a measure space. Let $H>0$ be integrable. Let $(E,d)$ be a separable metric space and $f,g: X \rightarrow E$ measurable functions. Let $d_H(f,g):= \int \min\{d(f(x),g(x)),1\}H(x)d\mu(x)$. Prove $f_n$ converges in measure to $f$ $\iff$ $\lim d_H(f_n, f) = 0$.

I think I have the "$\implies$" part (the sentence in bold later on is the part I am stuck on):

Let $1>\epsilon>0$ and $\delta>0$ and let $c:= \int H(x) < \infty$.

$d_H(f_n, f) = \int_{\{x|d(f_n(x),f(x)>\epsilon\}} \min\{d(f_n(x),f(x)),1\}H(x)d\mu(x) + \int_{\{x|d(f_n(x),f(x)\leq\epsilon\}} \min\{d(f_n(x),f(x)),1\}H(x)d\mu(x)$

From monotonicity and from our assumption that $f_n$ converges in measure to $f$, for a large enough $n$ we can bound the first term by:

$$\leq \int_{\{x|d(f_n(x),f(x))>\epsilon\}} H(x)d\mu(x) \leq c\mu(\{x|d(f_n(x),f(x))>\epsilon\} \leq c\delta$$

For the second term, since $\epsilon<1$, we obtain:

$$ \leq \int_{\{x|d(f_n(x),f(x)\leq \epsilon\}} \epsilon H(x)d\mu(x) \leq \epsilon c$$

Since these two hold for all $1>\epsilon>0$ and $\delta>0$ for a sufficiently large $n$, we obtain the conclusion.

For the "$\Longleftarrow$" part I am stuck with $H(x)$:

Let $1>\epsilon>0$ and $\delta>0$.

$$ \int_{\{x|d(f_n(x),f(x))> \epsilon \}} \epsilon H(x) \leq \int_{\{x|d(f_n(x),f(x))> \epsilon\}} \min\{1,d(f_n(x),f(x))\}H(x) \leq \int \min\{1,d(f_n(x),f(x))\}H(x)$$

Without the $H(x)$ I would be able to write:

$$\epsilon \mu(\{x|d(f_n(x),f(x))> \epsilon\}) \leq \int_{\{x|d(f_n(x),f(x))> \epsilon \}} \epsilon \leq \int_{\{x|d(f_n(x),f(x))> \epsilon\}} \min\{1,d(f_n(x),f(x))\} \leq \int \min\{1,d(f_n(x),f(x))\}$$

and then by taking the limit, the right-hand-side would tend to 0 and so for a large enough $n$ the left-hand-side is less than $\epsilon\delta$ and the conclusion follows. How can I do this with $H(x)$? Also - if there is feedback for improvement of the first part - would be great to know.

Upon further inspection, I don't think the statement as is holds. Take $X=\mathbb{R}$, $H(x)=\min\{ \frac{1}{x^2},1\}$, $f\equiv 0$ and $f_n=\mathbf{1}_{[n,n+1]}$.

Using classic definitions of measure convergence, we have that $d_H(f_n,f)\to 0$ but $f_n$ does not converge in measure to $f$.

I suspect you need more conditions on $(X,\mathcal{F},\mu)$ or $H$ for this to be true.