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Let $f: [0,1] \rightarrow \mathbb{R} $ a continuous function and $f > 0$. Let $F(x) = \int_0^x f(t)dt $ and let $F(1)=A$.

1. Show that $F$ is bijective function from $ [0,1]$ to $[0,A]$.

2. Show that for all $n \in \mathbb{N}^* $, there is a unique subdivision: $x_0 = 0 < x_1 < ... < x_{x-1} < x_n = 1 $, such that for $ k=0,1,..., n-1 $, we have: $\int_{x_{k}}^{x_{k+1}} f(t)dt = \frac{A}{n}$.

3. Show that $F^{-1} (\frac{kA}{n}) = x_{k+1}$ and $\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(x_{k}) = \frac{1}{A} \int_{0}^{1}(f(x))^2 dx. $

My work.

1. To show $F$ is bijective, I need to show it is injective and surjective, which I don't know how to do for this function.

2. $ A = \int_{0}^1f(x)dx = \sum_{k=0}^{n-1} \int_{x_{k}}^{x_{k+1}} f(x)dx $.

I don't see how to proceed to get $\int_{x_{k}}^{x_{k+1}} f(t)dt = \frac{A}{n}$.

3. $F$ is bijective, so it has an inverse function.

From question 2, $F^{-1}(\frac{kA}{n}) = F^{-1}\big(k \int_{x_{k}}^{x_{k+1}} f(t)dt \big)$ I don't see how to get $x_{k+1}$ from that.

For the limit, I tried to use Riemann sums but I got stuck too. Thank you for your help.

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For (1), note that $F$ is continuous and moreover $F'(x)=f(x)>0$ so $F$ is also strictly monotone. Hence $F$ is injective and by the Intermediate Value Theorem $F$ attains all the values between $F(0)=0$ and $F(1)=A$.

As regards (2), by the bijectivity for any $0\leq k\leq n$ there is a unique $x_k\in [0,1]$ such that $F(x_k)=\frac{kA}{n}$. Hence $$\int_{x_{k}}^{x_{k+1}} f(t)dt =F(x_{k+1})-F(x_k)=\frac{(k+1)A}{n}-\frac{kA}{n}= \frac{A}{n}.$$

Finally for (3), we have use Riemann sums, $$\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(x_{k}) =\lim_{n \rightarrow + \infty} \frac{1}{n} \sum_{k=1}^{n} f(F^{-1}(kA/n))\\=\frac{1}{A} \int_{0}^{A}(f(F^{-1}(t))) dt =\frac{1}{A} \int_{0}^{1}(f(x))^2 dx.$$ where $t=F(x)$, $dt=f(x) dx$.

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  • $\begingroup$ That proves it is injective. What about surjective? Thank you. $\endgroup$ Jul 12 '17 at 15:17
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    $\begingroup$ @Jacon See my edited answer. $\endgroup$
    – Robert Z
    Jul 12 '17 at 15:19
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    $\begingroup$ $f$ is bounded and integrable which implies that the integral function $F$ is continuous. $\endgroup$
    – Robert Z
    Jul 12 '17 at 15:30
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    $\begingroup$ For each $k$ we take $\frac{kA}{n}\in[0,A]$ and find $x_k=F^{-1}(\frac{kA}{n})$. Such $x_k$ exists by bijectivity of $F$ and $x_k<x_{k+1}$ because $F$ is strictly monotone $\endgroup$
    – Robert Z
    Jul 12 '17 at 15:42
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    $\begingroup$ For the RS the division points are $kA/n$ of the interval $[0,A]$. You are also correct but in the final step $dx$ becomes $dy/A$ (not $Ady$). For the final limit make the substitution $t=F(x)$. Hence $dt=F'(x)dx=f(x)dx.$ $\endgroup$
    – Robert Z
    Jul 12 '17 at 16:39

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