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What is an example of a sequence of continuous functions $f_n$ defined on the interval $[0,1]$ where $f_n \rightarrow 0$ pointwise and $\forall n, \int_0^1 f_n(x) dx = 1$?

I've thought of $f_n(x) = nx^n$, but for $x = 1$, $\lim_{n \rightarrow \infty} f_n(1) = +\infty$. I feel like I'm over thinking it and it's something dead simple.

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up vote 3 down vote accepted

Let $f_n(x) = \begin{cases} 0 & x \in [0,1-\frac{1}{n}) \\ 4n^2(x-(1-\frac{1}{n})) & x \in [1-\frac{1}{n},1-\frac{1}{2n}) \\ 4n^2(1-x) & x \in [1-\frac{1}{2n}, 1] \end{cases}$. (Basically a suitably large triangle with base $[1-\frac{1}{n},1]$.)

Then $f_n$ is continuous, $f_n(x) \to 0$ for all $x \in [0,1]$, and $\int f_n = 1$.

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Hrm, does it converge pointwise? If you take $x=1$, wouldn't we have $\lim_{n \rightarrow \infty} 4n^2$ which diverges? –  user52464 Dec 10 '12 at 5:54
    
First, if $x<1$, then eventually $x < 1-\frac{1}{n}$, and then $f_n(x) = 0$. Second, for all $n$, $f_n(1) = 0$, so the limit must be zero. I'm not sure how you get $f_n(1) = 4n^2$? –  copper.hat Dec 10 '12 at 5:56
    
$f_n$ is basically defined by joining the points $(0,0), (1-\frac{1}{n},0), (1-\frac{1}{2n},2n), (1,0)$. –  copper.hat Dec 10 '12 at 5:58
    
Woops, nevermind! –  user52464 Dec 10 '12 at 6:02
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