Why is $\lim_{n\rightarrow\infty}\int f_n \mathrm{d} \mu=\int f \mathrm{d} \mu$ not true We know that if $f_n\in \mathcal{L}^1(\mu)$, and $f_n \rightarrow f$ pointwise, it is not necessary true that $\lim_{n\rightarrow\infty} \int f_n \mathrm{d} \mu=\int f \mathrm{d} \mu$.
Some theorems restrict the property of the function sequence to make it work (dominated convergence theorem), my question is what is the essential reason why it is not true?
 A: Terry Tao's blog post on modes of convergence may be helpful. Specifically, he identifies three possible problems with $L^1$ convergence failing even when we have pointwise convergence. They are:


*

*Escape to Horizontal Infinity (Canonical: $f_n = \chi_{[n,n+1]}$)

*Escape to Width Infinity (Canonical: $f_n = \frac{1}{n} \chi_{[0,n]}$)

*Escape to Vertical Infinity (Canonical: $f_n = n \chi_{[0,1/n]} $)


This is in essence a classification of counterexamples. Yet we can think of them specifically as simple functions (since they approximate integrals by definition) and in turn as the canonical counterexamples.
Thinking about $L^1$ convergence in these terms help you see why theorems like the Dominated Convergence Theorem allow you to pass the limit through. No function can slide off to infinity as in (1), get fat enough, or (3) grow too large if there is a dominating function.
If you want to see if these are the only possible problems, consider a fourth option that fails. It doesn't escape to horizontal infinity or width infinity so most of the "mass" of $f_n$ must be in a bounded set. That is, there is an $N$ such that $f_n \chi_{[-N,N]^c} < \epsilon$. Neither does it escape to vertical infinity, so $f_n^{-1}((M,\infty))$ must be some $\epsilon$ of a set. Such a sequence of functions will have to converge in $L^1$ norm. I am not being precise here, but I hope you get the point.
A: Consider the sequence of functions $f_n = n\chi_{(1-\frac{1}{n},1)}$
You have $f_n\to 0$ pointwise almost everywhere, however $\int f_n d\mu= 1$ for all $n$
A: Besides of course having many counterexamples, the true reason is that to define a complete metric space we need more than just a notion of pointwise convergence. The theorems that make what you ask true, use precisely something that exploits the completeness of $L^1$.
