a question on dominated convergence theorem The theorem says that:
Assume that $f_n\rightarrow f$ a.e., with $f_n$ integrable for all n and $g$ is an integrable function such that $| f_n | \le g$. Then f is integrable and $\int \mathrm f\ \mathrm d\mu = \lim_n \int \mathrm f_n\ \mathrm d\mu.$
My question is if the following holds:
Assume that $f_n→f$ almost everywhere, with $f_n$ is bounded for all n. Then f is integrable and $\int f\mathrm f\ \mathrm d\mu = \lim_n \int \mathrm f_n\ \mathrm d\mu.$
Thanks.
 A: Simply being bounded does not imply being integrable. The function $f(x) \equiv 1$, for example, is bounded, but not integrable on the real line. 
A better counterexample shows that being bounded and integrable is also insufficient. Consider $f_n(x) = 1_{|x| < n}$, the characteristic function on $|x|<n$, so that $f_n(x) = 1$ if $|x|< n$ and $0$ otherwise.
Then for all $n$, $f_n$ is both bounded and integrable. But the limit is $f(x) \equiv 1$ as we mentioned above, not integrable. 
A: mixedmath's example shows that if the $f_n$ are all bounded and integrable and have a pointwise limit $f$, it still doesn't follow that $f$ is integrable.
Here are some examples to show that it is also possible for the $f_n$ to be bounded and integrable, with a pointwise limit $f$ that is also integrable, but still not to have $\int f = \lim_n \int f_n$.


*

*Let $f_n = n\cdot 1_{0\leq x\leq 1/n}$. Then each $f_n$ is integrable, and bounded (although the $f_n$'s are not uniformly bounded), and $f_n\to f=0$ almost everywhere. However, $\lim_n \int f_n = 1$ since each $\int f_n=1$, so $\lim_n \int f_n \neq \int f$.

*Let $f_n = 1_{n\leq x\leq n+1}$. Then again each $f_n$ is integrable, and bounded (this time even uniformly), and $f_n\to f=0$ everywhere. But again, each $\int f_n = 1$ while $\int f = 0$.
These two examples illustrate the ways in which a sequence of integrable functions can have a pointwise limit but the limit of their integrals is different than the integral of the pointwise limit. The mass can escape "upward" (first example) or "outward" (second example).
The Bounded Convergence Theorem works by insisting the $f_n$ be uniformly bounded (preventing the mass from escaping "upward") and only being applicable on a finite measure space (preventing the mass from escaping "outward").
The Dominated Convergence Theorem works because the dominator $g$ "traps" the $f_n$'s, preventing both effects.
