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 $\chi_{[2,3]}(1.5)=0$.
Let $f_n: X \to [0, \infty]$ be $\mu$-measurable functions, such that $f_n \leq f_{n+1}$ for each $n \in \mathbb N$ and assume that the sequence $\{ f_n \}_{n=1}^\infty$ converges to a function $f\colon X \to [0, \infty]$ in measure. By a theorem of F. Riesz we know, that that there exists a subsequence $\{ f_{n_j} \}_{j=1}^\infty$ of $\{ f_n \}_{n=1}^\infty$, such that $$ f_{n_j} \to f \quad \text{$\mu$-a.e.}$$
On the other hand, since $f_n \leq f_{n+1}$, we know that the limit $g := \lim_{n \to \infty} f_n$ exists, so we have $f = g$ $\mu$-a.e., i.e. $$f_n \to f \quad \text{$\mu$-a.e.}$$
Now the conditions for the monotone convergence theorem are fullfilled, and we deduce that
$$ \lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu \; . $$
Best Answer
The monotone convergence theorem requires $f_1(x) \leq f_2(x) \leq \cdots$ for all $x$.
But $$f_1(x) = \begin{cases} 1 & \text{if } x \in [0, 1], \\ 0 & \text{otherwise,} \end{cases}$$
and $$f_2(x) = \begin{cases}\frac12 & \text{if } x \in [0, 2], \\ 0 & \text{otherwise}\end{cases}$$
Notice that $f_1(0) = 1 \not\leq \frac12 = f_2(0)$. So the hypothesis fails.