Here's an example of a sequence, mentioned by Sam in the comments, of a sequence that converges pointwise nowhere on $[0,1]$ but $\int_{[0,1]} f_n\rightarrow 0$:
Let
$\ \ f_1=\chi_{[0,1]}$, $f_2=\chi_{[0,{1\over2}]}$, $f_3=\chi_{[{1\over2},1]}$, $f_4=\chi_{[0,{1\over4}]}$, $f_5=\chi_{[{1\over4},{2\over4}]}$, $f_6=\chi_{[{2\over4},{3\over4}]}$, $f_7=\chi_{[{3\over4},1]}$, $\ldots$.
Here $\chi_A$ is the function whose value is 1 on the set $A$ and $0$ on the set $A^C$.
First, for the "almost-everywhere" MCT:
Suppose that $f_n,\ f:X\rightarrow\mathbb{R}$ are such that $f_n\nearrow f$ a.e., where $(X,\mathcal{A},\mu)$ is a measure space.
This means that there is a measureable set $E\subseteq X$ such that $\mu(E)=0$, and we have that $f_n(x)\nearrow f(x)$ for all $x\in X\setminus E$.
If we define new functions $g_n,g:X\rightarrow\mathbb{R}$ by
$$
g_n(x)=\begin{cases}f_n(x) & \text{if }x\notin E\\0 & \text{if }x\in E\end{cases},\qquad g(x)=\begin{cases}f(x) & \text{if }x\notin E\\0 & \text{if }x\in E\end{cases}.
$$
Why should we care about these functions? You can prove:
- $\int g_n\,d\mu=\int f_n\,d\mu$ for all $n$, since these functions differ on a set of measure $0$; similarly, $\int g\,d\mu=\int f\,d\mu$.
- For every $x\in X$ (not just $x\in E$), $g_n(x)\nearrow g(x)$. Thus, by the MCT, $\int g_n\,d\mu\rightarrow\int g\,d\mu$.
Combining these, we see that $\int f_n\,d\mu\rightarrow\int f\,d\mu$, even though we relaxed our monotonicity assumption a bit.
Now, for his counterexample if we completely remove the monotonicity condition: here we have $f_n:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f_n=\chi_{[n,n+1]}$.
You asked why $f_n$ converges pointwise? Pick any $x\in \mathbb{R}$. Note that if we let $n\rightarrow\infty$, then eventually $n>x$; but if $n>x$, then $\chi_{[n,n+1]}(x)=0$. This holds for all $n>x$, so that $\chi_{[n,n+1]}(x)\rightarrow 0$ as $n\rightarrow\infty$.
This holds for every $x\in\mathbb{R}$; so, the sequence $\{\chi_{[n,n+1]}\}$ converges pointwise to the $0$-function.
This stands in contrast to the fact that, as your instructor said, $\int\chi_{[n,n+1]}\,d\mu=\mu([n,n+1])=1$ when we take $\mu$ to be Lebesgue measure.
Why does this not violate the MCT? Because the sequence isn't monotone increasing! Take, for instance, the fact that $\chi_{[1,2]}(1.5)=1$, but $\chi_{[2,3]}(1.5)=0$.
Best Answer
This is similar to the fact that pointwise convergence does not imply $L^1$ convergence.
Construct a sequence $f_n$, each with integral $1$, that converges to $0$ pointwise.