Measurable function satisfying $| \int_{E} fdx | \leq 1 $ for every Lebesgue-measurable set $E$ with $m(E) \leq 1 $ satisfies certain condition 
Let $f: \mathbb{R} \to \mathbb{R}$ be a measurable function satisfying $| \int_{E} fdx | \leq 1 $ for every Borel set $E$ with $m(E) \leq 1 $.
Prove that $$\lim_{n \to \infty} n m(\lbrace x \in \mathbb{R} : |f(x)| \geq n \rbrace) = 0.$$

$m$ is the standard Lebesgue measure. So this would be easy if the function were integrable, i.e. if $\int_{\mathbb{R}} fdx < +\infty$, but this doesn't have to be the case, since for example $f(x) = \frac{1}{2}$ satisfies this condition.
I've tried assuming the opposite, that there exists an $\varepsilon > 0$ and a subsequence $n_{k}$ such that $n_{k} m(\lbrace x \in \mathbb{R} : |f(x)| \geq n_{k} \rbrace) \geq \varepsilon$ for all $k \in \mathbb{N}$. What I know is that this sequence of sets, $\{A_{k}\}_{k=1}^{\infty}$, $A_{k} = n_{k} m(\lbrace x \in \mathbb{R} : |f(x)| \geq n_{k} \rbrace) \geq \varepsilon$, is a decreasing (i.e. $A_{k+1} \subseteq A_{k}$) sequence, and that its limit is the empty set, so it would be nice if I could prove that $m(A_{k}) < +\infty$ for some $k$ so that I could apply continuity from below and conclude that $m(A_{k}) \to 0$ when $k \to \infty$, but I can't even prove that.
Also, I have no idea how to construct a contradiction using the given condition.
Edit: Here's what I have so far:
$$(\forall \alpha \in \mathbb{R})(\forall k \in \mathbb{N}) | \int_{\alpha}^{\alpha+k} fdx|<k;$$
From the opposite assumption, we have $$\int_{A_{k}} |f|dx \geq \varepsilon$$ for all $k \in \mathbb{N}$.
 A: I assume $f$ is Borel measurable. If it is merely Lebesgue measurable, it is equal a.e. to a Borel measurable function, so this is without loss of generality.
Consider the set
$$F=\{x\mid f(x)\geq 2\}$$
If $m(F)> 1$ then $m(F\cap [-r,r])=1$ for some $r>0$ since this is continuous in $r$ and tends to $m(F)$ as $r\to\infty.$ But then $|\int_{F\cap [-r,r]}f|\geq 2$ which would contradict the assumption. So $m(F)\leq 1.$ By assumption, $\int 1_{f\geq 2} f=\int_F f\leq 1.$ (Here $1_X$ denotes the indicator function of the set described by $X$.)
Similarly $\int 1_{f\leq -2}|f|\leq 1.$
So for any $n\geq 2$ we have $\int 1_{|f|\geq n}|f|\leq 2.$ This integral tends to zero as $n\to\infty$ by the $L^1$ dominated convergence theorem. We get
$$n m(\lbrace x \in \mathbb{R} : |f(x)| \geq n \rbrace) \leq \int 1_{|f|\geq n}|f|\to 0$$
as required.
A: It is sufficient to show the result for $f$ positive. Define for $r>1$:
  $$ E_r = \{ f \geq r\}$$
We have by hypothesis on $f$:
  $$ r m (E_r) \leq \int_{E_r} f \leq m(E_r)+1$$
i.e. that $m(E_r)\leq 1/(r-1)<+\infty$.
Suppose now that the conclusion is false. Then there is $\delta>0$ such that the set:
  $$ S_\delta = \{ r>1 : r m\{ f \geq r\} \geq \delta >0 \}$$
is unbounded. Let $r_1\in S_\delta$. Given $r_k\in S_\delta$ by measurability of $f$ and $\sigma$-additivity of the measure, we may find  $r_{k+1}\in S_\delta$ for which:
  $$ r_k m \{ r_k \leq f < r_{k+1}\} \geq \delta/2.$$
But then
  $$ +\infty \leq \sum_{k\geq 1} \delta/2 \leq 
\sum_k r_k m \{ r_k \leq f < r_{k+1}\} \leq \int_{E_{r_k}\setminus E_{r_{k+1}}} f \leq \int_{E_{r_1}}f \leq  1/(r_1-1)<+\infty$$
a contradiction.
