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**Media_Man** Laurent: Thank you for the suggestion. I believe the following sketch is much closer to a formal proof.

Define $\displaystyle f_n:[0,1]\to[0,1]$ as follows: Divide the interval $\displaystyle [0,1]$ into $\displaystyle n$ subintervals $\displaystyle \Delta_i$ of width $\displaystyle \frac1n$. For each of these subintervals, let $\displaystyle f_n(\Delta_i)$ be a random element of $\displaystyle [0,1]$. Now define $\displaystyle g$ by sorting these subintervals from shortest to tallest, $\displaystyle g_n(\Delta_{i-1})\leq g_n(\Delta_i)$ for all $\displaystyle i$. It should be plain that $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$.

1. Define $\displaystyle g$ as the limiting function of $\displaystyle g_n$, ($\displaystyle g=\lim_{n\to\infty} g_n$), which is $\displaystyle g(x)=x$. Of course, $\displaystyle \int_0^1 g(x)dx=\frac12$.

Define $\displaystyle f$ as the limiting function of $\displaystyle f_n$, ($\displaystyle f=\lim_{n\to\infty} f_n$)

2. Since $\displaystyle \int_0^1 f_n(x)dx=\int_0^1 g_n(x)dx$ for all $\displaystyle n$, $\displaystyle \int_0^1 f(x)dx=\int_0^1 g(x)dx=\frac12$.

Are steps 1 and 2 valid?