real analysis - Problems proving that if $f_nrightarrow f$ pointwise and $int_R f=lim_{n}int_R f_n$ then $int_E f=lim_{n}int_E f_n$ for meas $E subseteq R$.

This is a problem from Royden 4th edition (updated printing). Problem 4.22

Let $\{f_n\}$ be a sequence of nonnegative measurable functions on $\mathbb{R}$ that converges pointwise on $\mathbb{R}$ to $f$ and $f$ be integrable over $\mathbb{R}$. Show that
\begin{equation}
\text{if}~\int_\mathbb{R}f=\lim_{n\rightarrow \infty} \int_\mathbb{R} f_n,~\text{then}~\int_E{f}=\lim_{n\rightarrow \infty} \int_E f_n,~\text{for any measurable set $E$}.
\end{equation}

Solution so far...

I've been able to deduce that the problem statement follows if $\lim_{n\rightarrow \infty} \int_E f_n$ exists for any measurable set $E$.

Suppose that under the assumptions of the problem, $\lim_{n\rightarrow \infty} \int_E f_n$ exists for any measurable set $E$.

Proof that if $\int_\mathbb{R} f = \lim_{n\rightarrow\infty} \int_\mathbb{R} f_n$ then $\int_E f = \lim_{n\rightarrow\infty} \int_E f_n$. We prove by contradiction that equality holds.

Suppose equality does not hold. By
Fatou's Lemma we know that $\int_E f \leq \lim_{n\rightarrow\infty} \int_E f_n$, so since equality doesn't hold then $\int_E f < \lim_{n\rightarrow\infty} \int_E f_n$. Then it follows that
\begin{align*}
\int_\mathbb{R} f &= \int_E f + \int_{\mathbb{R} \sim E} f &(\text{additivity over domains})\\
&\leq \int_{E} f + \lim_{n\rightarrow \infty} \int_{\mathbb{R} \sim E} f_n & \text{(Fatou's)}\\

&< \lim_{n\rightarrow \infty} \int_E f_n + \lim_{n\rightarrow \infty} \int_{\mathbb{R}\sim E} f_n\\
&= \lim_{n\rightarrow \infty} \int_\mathbb{R} f_n
\end{align*}
which contradicts our assumption. Therefore $\int_E f = \lim_{n\rightarrow\infty} \int_E f_n$.

The part I'm having trouble with is the initial claim that under the problems assumptions $\lim_{n\rightarrow \infty} \int_E f_n$ exists for any measurable set $E$. Am I approaching this in a reasonable way? Any hints on how to prove this last bit?

Answer

$\int (f-f_n)^{+} \to 0$ by DCT because $(f-f_n)^{+} \leq f$ and $(f-f_n)^{+} \to 0$. Also $\int (f-f_n) \to 0$ by hypothesis. Subtract the first from the second to get $\int (f-f_n)^{-} \to 0$. Add this to $\int (f-f_n)^{+} \to 0$ to get $\int |f-f_n| \to 0$. For any measurable set $E$ we have $\int_E |f-f_n|\leq \int_{\mathbb R} |f-f_n| \to 0$ which implies $\int_E f_n \to \int_E f$.