Showing $f_n \rightarrow f$ in $L^p(\Bbb{R})$ Suppose we know the following
(i) $f_n$ converges to $f$ in measure. 
(ii) there is a finite measure set, $E$, such that for any $\epsilon >0$ there exists a $\delta>0$ such that if $m(E)< \delta$, then
$$\int_E \vert f_n \vert^p < \epsilon.$$
(iii) for any $\epsilon >0$ there exists a measurable set $E$ such that if $m(E) < \infty$, then
$$\int_{E^c} \vert f_n \vert^p < \epsilon$$
Show that $f_n \rightarrow f$ in $L^p(\Bbb{R})$. So I thought of writing
$$\int_{\Bbb{R}} \vert f_n - f \vert^p$$
As
$$\int_{E_\epsilon} \vert f_n - f \vert^p + \int_{E_\epsilon^c}\vert f_n - f \vert^p$$
Where $E_\epsilon =\{x : \vert f_n - f \vert < \epsilon\}$
Am I decomposing the integral the correct way? Any $\textbf{Hints}$ greatly appreciated. In this case, would the latter integral go to zero as $f_n$ converges to $f$ in measure?
 A: The second assumption should read without "there is a finite measure set, $E$, such that".
The third assumption should read "for any $\varepsilon >0$, there exists a measurable set $E$ of finite measure such that for each $n$, $\int_{E^c}\lvert f_n\rvert^p<\varepsilon$.
By an application of Fatou's lemma to an almost everywhere convergent sequence, we can see that the estimates in (ii) and (iii) also hold with $f_n$ replaced by $f$, hence we can assume without loss of generality that $f=0$.  Indeed, let $(f_{n_k})$ be an almost everywhere convergent subsequence to $f$. Then for $E$ such that $\mu(E)<\delta$, $$\int_E \lvert f\rvert^pd\mu=\int_E \liminf_{k\to\infty} \lvert f_{n_k}\rvert^pd\mu\leqslant \liminf_{k\to\infty}\int_E  \lvert f_{n_k}\rvert^pd\mu\leqslant\varepsilon$$
and the same argument shows that for the set $E$ of finite measure such that $\int_{E^c}\lvert f_n\rvert^p<\varepsilon$, $\int_{E^c}\lvert f\rvert^p\leqslant\varepsilon$. If we show the result in the particular case where $f=0$, the general case follows by noticing that if $(f_n)$ satisfies (i)-(ii)-(iii), so does $f_n-f$.
Pick a positive $a$ and $\varepsilon>0$ and let $E_{n,a}=\{x:\lvert f_n \rvert>a\}$. Let $\delta$ be such that
$$
\sup_{E:\mu(E)<\delta}\sup_{n\geqslant 1}\int_E \vert f_n \vert^p < \varepsilon.
$$
Notice that by (i), there exists $N$ such that for each $n\geqslant N$,  $\mu(E_{n,a})<\delta$ hence $\sup_{n\geqslant 1}\int_{E_{n,a}} \vert f_n \vert^p < \varepsilon.$
Let $A$ be a set of finite measure such that $\sup_{n\geqslant 1}\int_{A^c}\vert f_n \vert^p < \varepsilon.$ Then
$$
\int \lvert f_n\rvert^p\leqslant \int_{E_{n,a}}\lvert f_n\rvert^p
+\int_{E_{n,a}^c\cap A}\lvert f_n\rvert^p+\int_{E_{n,a}^c\cap A^c}\lvert f_n\rvert^p
\leqslant \varepsilon+ a^p\mu(A)+\varepsilon.
$$
Since $a$ is arbitrary, this gives $
\int \lvert f_n\rvert^p\leqslant 2\varepsilon$ and the conclusion follows.
