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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?)

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    I'm surprised that you know what $L^2$ is without knowing the monotone convergence theorem. Where did you learn this material from? By the way, what does "p.w." mean?2011-08-21
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    p.w. = pointwise I guess2011-08-21
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    But the pointwise limit of $L^2$ functions need not be $L^2.$ Consider the family given by $f_n(x) = \frac{1}{\sqrt{x+1/n}}.$ Then $f_n$ is a positive, nondecreasing sequence of $L^2([0,1])$ functions, but the pointwise limit ${1}/{\sqrt{x}}$ is not $L^2.$ Moreover, neither is $|f_n -f|$ for any $n.$ The monotone convergence theorem only answers the $L^1$ analog of mathfreaks question.2011-08-21

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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 $f\notin L^2$.

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    Further explanation: $|f_n-f|^2$ decreases to zero pointwise.2011-08-21
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    Why? It seems to me that the monotone convergence theorem assures that $f_n \to f$ in the $1$-norm. Why also in $2$-norm?2011-08-21
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    @GEdgar: sorry, I was working on that, but I was drawn away for a while.2011-08-21
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    Yes. But this only works if the pointwise limit is bounded.2011-08-21
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    @jspecter: I noticed that and appended my answer appropriately.2011-08-21
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    Monotone convergence requires both monotone increasingness and nonnegativity.2011-08-21
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    @Zarrax: I was using the version not requiring non-negativity, derived from the non-negative version: Let $f^+(x)=\left\{\begin{array}{cl}f(x)&\mathrm{if\ }f(x)\ge0\\0&\mathrm{if\ }f(x)<0\end{array}\right.$ and $f^-(x)=\left\{\begin{array}{cl}0&\mathrm{if\ }f(x)\ge0\\-f(x)&\mathrm{if\ }f(x)<0\end{array}\right.$ Then $\{f_n^+\}$ and $\{f_0^--f_n^-\}$ are non-negative, non-decreasing sequences of functions tending to $f^+$ and $f_0^--f^-$ respectively. Apply the non-negative restricted version to $f_n^++(f_0^--f_n^-)-f_0^-(=f_n)$ which converges to $f^++(f_0^--f^-)-f_0^-(=f)$2011-08-22
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    @Zarrax: the ideas of which are used in your answer, I see.2011-08-22
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    @robjohn so I guess another way of saying this is that in your answer you are applying the normal monotone convergence thm to $\int(|f|^2 - |f_n - f|^2)$ and then subtracting off $\int|f|^2$ from both sides.2011-08-23
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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.