Counter example for integral differing from limit of integrals On exercise $4.U$ of Bartle's Element of Integration, we are asked to give a counter example on a sequence $f_n$ of functions such that $\lim_{n\to\infty}f_n=f$, $\int f =\lim_{n\to\infty}\int f_n$
And If $\lim_{n\to\infty}\int f_n=\infty$ then  $\int_{E} f \neq\lim_{n\to\infty}\int_{E} f_n$
Where $E$ is a subset of my $\sigma-algebra$
I tried to think either of a function or a measure space but the simplest cases are not seeming to work, any suggestions?
In lebesgue dominated convergence theorem the place this hypothesis appears for me is so i can use reverse fattou's lema, so maybe such a function that does not apply to fattou's lema Will be the one
 A: A counterexample is $f_n(x) = n^2\chi_{(0,1/n]}(x)+1$, where we have $f_n \to 1$ pointwise on $(0,\infty)$ and 
$$\lim_{n \to \infty}\int_{(0,\infty)}f_n = \int_{(0,\infty)}f = +\infty,$$
but 
$$\lim_{n \to \infty}\int_{(0,1)}f_n = \lim_{n \to \infty}(n+ 1 - 1/n) = + \infty \neq \int_{(0,1)} f = 1$$
Stated precisely, the related theorem applies to a sequence $(f_n)$ of nonnegative functions on a measure space $X$ and where $f_n(x) \to f(x)$ and $\lim_{n \to \infty}\int_X f_n = \int_X f < +\infty$. In this case for any measurable set $E \subset X$, we have $\lim_{n \to \infty}\int_E f_n = \int_E f$.
The proof uses Fatou's lemma where 
$$\begin{align}\int_E f &= \int_E \liminf_{n \to \infty}f_n \\ &\leqslant \liminf_{n \to \infty}\int_E f_n \\ &\leqslant \limsup_{n \to \infty}\int_E f_n \\ &= \limsup_{n \to \infty}\left(\int_X f_n - \int_{X \setminus E} f_n \right)\\ &= \int_Xf - \liminf_{n \to \infty} \int_{X \setminus E} f_n \\ &\leqslant \int_Xf - \int_{X \setminus E}\liminf_{n \to \infty}  f_n \\ &= \int_Xf - \int_{X \setminus E}  f \\ &= \int_E f\end{align}$$
Thus, $\lim_{n \to \infty}\int_E f_n = \int_E f$. This proof breaks down,of course, if the integral of $f$ and the limit of integrals of $f_n$ over $X$ are not finite.
A: The question seems a little unclear, but using Mark's comment.  It is east to construct such a sequence $f_n(x)=n^2x$, for $0\le x\le \frac{1}{n}$ and $=0$ otherwise.  So $\int f_n(x)=1/2$.  However $f(x)=\lim_{n\to \infty} f_(x)=0$ for all $x$, so $\int f(x)=0$.
