Does pointwise monotone convergence imply 2-norm convergence? Let $\mu$ be the Lebesgue measure on the segment $[0,1]$. Let $(f_n)$ be a sequence in $L^{2}([0,1])$ such that $f_n\rightarrow f$  p.w. and $f_{n+1}\geq f_{n}$. Does $f_n\rightarrow f$ also in the 2-norm? (what if $(f_n)$ is a sequence of simple functions?)
 A: This is what is called the Monotone Convergence Theorem, and it is one of the most important convergence theorems for integration of functions and summation of series.
In your particular case, $f_n\to f$ pointwise and monotonically increasing, so $f_n-f\to 0$ pointwise and monotonically increasing.  This means that $|f_n-f|^2\to 0$ pointwise and  monotonically decreasing.  Thus, $\int_0^1|f_n(x)-f(x)|^2{\rm d}x\to 0$ by monotone convegence.  If you want to stick with monotonically increasing functions, $-|f_n-f|^2\to 0$ pointwise and monotonically increasing.
I just noticed that nowhere was it specified that $\|f_n\|_{L^2}$ was bounded.  If $\lim_{n\to\infty}\|f_n\|_{L^2}=\infty$, then $f\notin L^2$.
A: As long as the limit function $f$ is in $L^2$ this is true. Let $f_n^+$ and $f^+$ denote the positive parts of $f_n$ and $f$ respectively, and $f_n^-$ and $f^-$ the negative parts. Then by the monotonicity conditions you have
$$|f_n| \leq |f_n^+| + |f_n^-| \leq |f^+| + |f_1^-|$$
$$|f| \leq |f^+| + |f^-| \leq |f^+| + |f_1^-|$$
So if $g(x)$ denotes $|f^+| + |f_1^-|$, then $g(x)$ is an $L^2$ function that dominates $|f|$ and each $|f_n|$. Hence 
$$|f_n - f|^2 \leq (|f_n| + |f|)^2 \leq 4g^2$$
Since $4g^2$ is in $L^1$, the dominated convergence theorem gives 
$$\lim_{n \rightarrow \infty} \int_0^1 |f_n - f|^2 = 0$$
This is the desired $L^2$ convergence.
