Limit of Integral over Any Measurable Subset Exists

Question

Let $\{f_n\}$ be a sequence of nonnegative measurable functions on $\Bbb{R}$ that converges pointwise on $\Bbb{R}$ to $f$ and $f$ be integrable over $\Bbb{R}$. Show that $$\mbox{If } \int_\Bbb{R} f = \lim_{n \to \infty} \int_\Bbb{R} f_n, \mbox{ then } \int_E f = \lim_{n \to \infty} \int_E f_n$$ for any measurable set $E \subseteq \Bbb{R}$.

I found a solution to this problem here (see the third and fourth pages), but I don't understand the following line from the proof:

$$\int_{\Bbb{R} - E} f \le \liminf_{n} \int_{\Bbb{R} - E} f_n = \int_\Bbb{R}f - \limsup_n \int_E f_n$$

Is the equality valid? The only way I could see to justify this step was by noting that $\int_\Bbb{R} f_n = \int_E f_n + \int_{\Bbb{R} - E} f_n$ or $\int_{\Bbb{R} - E} f_n = \int_{\Bbb{R}} f_n - \int_E f_n$, and then

$$\liminf_{n} \int_{\Bbb{R} - E} f_n = \liminf_n (\int_{\Bbb{R}} f_n - \int_E f_n) = \int_\Bbb{R} f - \limsup_n \int_E f_n$$

But would this not be fallacious, as we do not know whether the $f_n$ are integrable and therefore cannot subtract those integrals, which are potentially infinite, as we see fit? Is this solution incorrect, or am I misunderstanding something?

Answer 1

$f_n$ is integrable for $n$ large enough, because we have the assumption $$\int f = \lim_{n\rightarrow\infty} \int f_n.$$ It works in the following way: The convergence gives you an $n_0$, such that for every $n\geq n_0$ $$\int f_n \leq 1+\int f < \infty.$$ Since $f_n$ is nonnegativ, it is integrable.

Answer 2

It seems the link to the proof from the original post no longer works. For the sake of having full context (and for those who are curious), here is a proof to the problem OP is referencing.

By assumption, $\int_{\mathbb{R}}^{}fd\mu=\lim_{n\to\infty}\int_{\mathbb{R}}^{}f_nd\mu$, which means that if $f$ is integrable, $f_n$ is also integrable for large enough $n$. (See answer by humanStampedist)

Next, let $E$ be any subset of $\mathbb{R}$. $\mathbb{R}$ can be decomposed into disjoint sets $\mathbb{R}=E\cup(\mathbb{R}\setminus E)$.

By Fatou's Lemma, we can claim \begin{align} \int_{\mathbb{R}\setminus E}^{}\liminf_nf_n&\leq\liminf_n\int_{\mathbb{R}\setminus E}^{}f_n\nonumber\\ \implies\int_{\mathbb{R}\setminus E}^{}f&\leq\liminf_n\int_{\mathbb{R}\setminus E}^{}f_n \end{align} where the second line follows from pointwise convergence of $f_n$ to $f$. Using integration over disjoint sets, we can write \begin{align} \liminf_n\int_{\mathbb{R}\setminus E}^{}f_n&=\liminf_n\int_{\mathbb{R}}^{}f_n-\limsup_n\int_{E}^{}f_n\nonumber\\ &=\int_{\mathbb{R}}^{}f-\limsup_n\int_{E}^{}f_n \end{align} where the second line follows from the assumption that $\int_{\mathbb{R}}^{}fd\mu=\lim_{n\to\infty}\int_{\mathbb{R}}^{}f_nd\mu$ Rearranging the above results yields \begin{align} \int_{\mathbb{R}\setminus E}^{}f&\leq\int_{\mathbb{R}}^{}f-\limsup_n\int_{E}^{}f_n\nonumber\\ \implies\limsup_n\int_{E}^{}f_n&\leq\int_{\mathbb{R}}^{}f-\int_{\mathbb{R}\setminus E}^{}f\nonumber\\ \implies\limsup_n\int_{E}^{}f_n&\leq\int_{E}^{}f \end{align} By again using Fatou's Lemma, we know \begin{equation} \int_{E}^{}f\leq\liminf_n\int_{E}^{}f_n \end{equation} The only way for $\limsup_n\int_{E}^{}f_n\leq\int_{E}^{}f$ and $\int_{E}^{}f\leq\liminf_n\int_{E}^{}f_n$ to simultaneously be true is if we have equality to the limit. Thus \begin{equation}\label{eqn:P2a5} \lim_n\int_{E}^{}f_n=\int_{E}^{}f \end{equation}