# Problems proving that if $f_n\rightarrow f$ 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 $$$$\text{if}~\int_\mathbb{R}f=\lim_{n\rightarrow \infty} \int_\mathbb{R} f_n,~\text{then}~\int_E{f}=\lim_{n\rightarrow \infty} \int_E f_n,~\text{for any measurable set E}.$$$$

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?

$$\int (f-f_n)^{+} \to 0$$ 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 $$\int_E f_n \to \int_E f$$.
• I cannot think of a way of proving the existence of $\lim \int_E f_n$ for each $E$ by any other method. So, as far as I can see, your argument cannot be completed. – Kavi Rama Murthy Oct 19 '18 at 4:58