It seems that we can get a stronger result that there exists a positive measurable function $h$ such that there is a measurable set $A \subseteq \mathbb{R}$ such that $m(\mathbb{R} \setminus A) = 0$ and $f_n h \to 0$ uniformly on $A$. Note that this clearly implies that $f_n h \to 0$ in measure.
Denote by $m$ the standard Lebesgue measure on $\mathbb{R}$ and take any expression $\mathbb{R} = \cup_{i=1}^{\infty} E_i$ where each $E_i$ is measurable and $m(E_i) < \infty$ (for example by bounded intervals $E_i$'s). For each $j \geq 1$, we have, by the basic standard theorem of Egorov (https://en.wikipedia.org/wiki/Egorov%27s_theorem), that there exists a measurable subset $A_{ij} \subseteq E_i$ with $m(E_i \setminus A_{ij}) < 1/j$ and $f_n \to 0$ uniformly on $A_{ij}$. Then note that $m(E_i \setminus \cup_{j=1}^{\infty} A_{ij}) = 0$. Taking $A = \cup_{i,j \geq 1} A_{ij}$, we have also that $\displaystyle \mathbb{R} \setminus A = (\bigcup_i E_i )\setminus A \subseteq \bigcup_{i \geq 1} (E_i \setminus \bigcup_{j \geq 1} A_{ij})$, so that $m(\mathbb{R} \setminus A) = 0$. Since $\{ A_{ij} \}_{i,j \geq 1}$ is a countable family of sets, we can reindex this so that $A = \cup_{i=1}^{\infty} A_i$ where each $A_i$ is measurable and $f_n \to 0$ uniformly on each $A_i$.
Set $B_k = \{ x \, | \, \sup_n |f_n(x)| \leq k \}$. Since $f_n \to 0$ pointwise on $\mathbb{R}$, for each $x \in \mathbb{R}$ we must have $\sup_n |f_n(x)| < \infty$ (why?), so that $\mathbb{R} = \cup_{k=1}^{\infty} B_k$.
Now take $\displaystyle h = \sum_{i,j \geq 1} \frac{1}{j2^{ij}} \chi_{A_i \cap B_j} + \chi_{\mathbb{R} \setminus A}$, where as usual $\chi_E$ denotes the indicator function on $E$, i.e. $1$ on $E$ and $0$ elsewhere. Note that $\displaystyle \sum_{i,j \geq 1} \frac{1}{j 2^{ij}} \leq \sum_{i,j} \frac{1}{2^{ij}} = \sum_i \sum_j \frac{1}{2^{ij}} = \sum_i \frac{1}{2^i - 1} \leq \sum_i \frac{1}{2^{i-1}} = 2 < \infty$ and that $h > 0$ everywhere on $\mathbb{R}$, so $h$ is a positive (real-valued) measurable function on $\mathbb{R}$.
Given $\epsilon > 0$, we can take some large positive integer $s$ so that $\displaystyle \frac{4}{2^s - 1} < \frac{\epsilon}{2}$. With this choice, we have
$$\sum_{i \geq 1} \sum_{j \geq s} \frac{1}{2^{ij}} = \sum_{i=1}^{\infty} \frac{1}{2^{is} - 2^{is - i}} = \sum_{i \geq 1} \frac{1}{2^{is - i}(2^i - 1)} \leq \sum_{i \geq 1} \frac{1}{2^{is -i}(2^{i-1})} = \sum_{i \geq 1} \frac{1}{2^{is - 1}} = 2 \sum_{i \geq 1} \frac{1}{2^{is}} = \frac{2}{2^s - 1} < \frac{\epsilon}{4}.$$
Now,
$\displaystyle |f_n h| \leq \sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{i \geq s, j \geq 1} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{j \geq s, i \geq } \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + |f_n| \chi_{\mathbb{R} \setminus A} \leq \sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{i \geq s, j \geq 1} \frac{1}{2^{ij}} \chi_{A_i \cap B_j} + \sum_{j \geq s, i \geq 1} \frac{1}{2^{ij}} \chi_{A_i \cap B_j} + |f_n| \chi_{\mathbb{R} \setminus A}.$
In the very latter expression, the middle two terms sum to something smaller than $\frac{\epsilon}{2}$, and in the first term, $\sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j}$, we have uniform convergence to $0$ (it's a finite sum of sequences of functions that converge uniformly to $0$), so there exists $N$ such that if $n \geq N$ then $\sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} < \epsilon / 2$. So that for $n \geq N$, we have $|f_n(x) h(x)| < \epsilon$ for all $x \in A$. So $f_n h \to 0$ uniformly on $A$.
Best Answer
In the backwards direction you don't have uniform convergence, only that $\int_E \min(1,|f_n-f|)\,dm \to 0$. Also, the conclusion $$\int_E \min(1,|f_n-f|)\,dm \to 0 \implies \int_E |f_n-f|\,dm \to 0$$ is not entirely spelled out in your question.
Anyway, for $0 < \varepsilon < 1$ you can use Markov's inequality to obtain $$m(|f_n-f| > \varepsilon) =m\Big(\min(1,|f_n-f|) > \varepsilon\Big)\le \frac1\varepsilon \int_E\min(1,|f_n-f|)\,dm \xrightarrow{n\to\infty} 0.$$
Conversely, assume that $f_n \to f$ in measure $m$ and let $0 < \varepsilon <1$. Pick $n_0 \in \Bbb{N}$ such that for $n \ge n_0$ we have $$m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right) < \frac\varepsilon2.$$ Then for all $n \ge n_0$ we have \begin{align} \int_E \min(1,|f_n-f|)\,dm &\le \int_{|f_n-f| \le \frac\varepsilon{2m(E)}} |f_n-f|\,dm + \int_{|f_n-f| > \frac\varepsilon{2m(E)}} 1\,dm\\ &\le \frac\varepsilon{2m(E)}m\left(|f_n-f| \le\frac\varepsilon{2m(E)}\right) + m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right)\\ &\le \frac\varepsilon2 + \frac\varepsilon2\\ &=\varepsilon \end{align} so we conclude $\int_E \min(1,|f_n-f|)\,dm \to 0$.