Here is my attempt:
Define $f_n=\chi_{E\Delta(E+ \frac{1}{n})}$. Then $f_n$ decreses with regard to $n$. Since $$m(E\Delta(E+\frac{1}{n})) = \int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm,$$ it suffices to show $$lim_{n\to\infty}\int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm = 0.$$ According to Lebesgue dominated convergence theorem, $$lim_{n\to\infty}\int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm = \int_\mathbb{R}\lim_{n\to\infty}\chi_{E\Delta(E+ \frac{1}{n})}dm.$$ Thus we only need to show $$\chi_{E\Delta(E+ \frac{1}{n})}\overset{a.e.}\to0.$$
EDIT:
By the Approximation Theorem of Measure Theory, $\forall \epsilon > 0$ there exist a finite number of disjoint intervals $\{I_k\}_{k=1}^N$ such that $m(E\Delta(\cup_{k=1}^NI_k)) < \epsilon$. Assume $F = \cup_{k=1}^NI_k$, then $m(E\Delta F) < \epsilon$. Define $f_n = \chi_{E\Delta (E+1/n)}$, $g_n = \chi_{F\Delta (F+1/n)}$.
Step 1. I will show $\int \mid f_n – g_n\mid dm < 2\epsilon$. Since $$(E\Delta (E+1/n))\Delta (F\Delta (F+1/n))\subseteq (F\Delta E)\cup ((F + 1/n)\Delta (E + 1/n))$$ we have $$\int \mid f_n – g_n\mid dm = m((E\Delta (E+1/n))\Delta (F\Delta (F+1/n))) \leq m(F\Delta E) + m((F + 1/n)\Delta (E + 1/n)) < 2\epsilon$$
Step 2. I will show $\lim_{n\to \infty}\int \mid g_n\mid dm = 0$. $\{I_k\}_{k=1}^N$ can be written as $\{[a_k,b_k)\}_{k=1}^N$, then $$\int \chi_{F\Delta (F+1/n)}dm = m(\cup_{i=1}^N([a_i,a_i+1/n)\cup [b_i,b_i+1/n)))\leq \frac{2}{n}N$$ Therefore $$\lim_{n\to \infty}\mid g_n\mid dm = \lim_{n\to \infty} g_n dm = 0$$
Step 3. $$\int \mid f_n – g_n\mid dm < 2\epsilon$$ $$\implies \int \mid f_n\mid dm – \int \mid g_n\mid dm < 2\epsilon$$ $$\implies \lim_{n\to \infty}\int \mid f_n \mid dm < 2\epsilon$$ Let $\epsilon \to 0$, we get $\lim_{n\to \infty}\int\mid f_n\mid dm=0$. Since $f_n$ is non-negative, $\lim_{n\to \infty}\int f_n dm=0$
Best Answer
It is neither true that $f_n $ is decreasing nor is it true that $f_n \to 0$a.e.
By the Approximation Theorem of Measure Theory (Ref. Halmos's book) we can find a finite disjoint union $F$ of intervals of the type $[a_i,b_i), 1 \leq i \leq N$ such that $m (E\Delta F) <\epsilon$. Let $g_n= \chi_{F\Delta (F+\frac 1 n)}$. I will let you verify that $\int |f_n-g_n| <2 \epsilon$ and $\int g_n \leq \frac 2 n N \to 0$.