Uniformly integrable sequence of functions on R pointwise to a not integrable function I have to prove that the following property (Royden 4th edition page 93):
Assume $E$ has finite measure. Let the sequence of functions $(f_n)$ be uniformly
integrable over $E$. If $\{f_n\} \rightarrow f$ pointwise a.e. on $E$, then $f$ is integrable over $E$.
is false if $E = \mathbb{R}$
I understand the proof from the book for $E$ finite measure, but any help on how to prove the property is false if  $E = \mathbb{R}$. Or do I have to find an example? Thanks. 
 A: We work with the domain $\mathbb{R}$, define
$$f_n (x) = \chi_{[0,n)}(x)$$
we have $f_n \rightarrow \chi_{[0,\infty)}(x)$ pointwise a.e. and $\{f_n\}$ is uniformly integrable. But $\chi_{[0,\infty)}(x)$is not integrable.
A: (1) $f_n \to \chi_{[0,\infty)}$ pointwise almost everywhere.
Take any $x \in \mathbb{R}$. If $x < 0$, then $f_n(x) = 0$ for all $n$ and $\chi_{[0,\infty)}(x) = 0$. If $x \ge 0$, then $f_n(x) = 1$ for all $n > x$ and $\chi_{[0,\infty)}(x) = 1$.
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2). $(f_n)_n$ is uniformly integrable over $\mathbb{R}$.
Take $\epsilon > 0$. We wish to show that there is some $\delta > 0$ such that $|E| < \delta$ implies $\int_E |f_n(x)|dx < \epsilon$ for any $n \ge 1$. But since $|f_n(x)| \le 1$ for any $n \ge 1$ and $x \in \mathbb{R}$, taking $\delta = \epsilon$ suffices (since $\int_E |f_n(x)|dx \le \int_E 1dx = |E| < \delta$). 
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3). $\chi_{[0,\infty)}$ is not integrable over $\mathbb{R}$. 
To show this, you can use monotone convergence theorem. Since the $f_n$'s increase pointwise to $\chi_{[0,\infty)}$, we know $n = \int f_n \to \int \chi_{[0,\infty)}$, implying $\int \chi_{[0,\infty)} = \infty$.
A: Consider $$g_n (x) = \cases{1 , \, \,\text {if} \,\,\, x = r_1, r_2, \ldots , \,\,\text {or}\,\,r_n\\ 0 , \,\,\, \text{otherwise}}$$
where $\{r_n\}$ is a sequence that is one-to-one correspondence with $\mathbb N$ and range $\mathbb Q$, the sequence $\{g_n\}$ converges to the characteristic function of the set $\mathbb Q$ restricted to $[0,1]$. 
