We have $(X_i)_{i \in \mathbb Z}$ iid random variables with $1\le X_i \le2$ almost surely.
We define $X(x,\omega) \equiv X_i (\omega)$ if $x\in [i,i+1[$ (so it is bounded almost surely) and, for $\epsilon >0$, $X_\epsilon (x, \omega) \equiv X(x/\epsilon, \omega)$.
Define for $x\in [0,1]$
$$u_\epsilon(x,\omega)= \int_0^x \frac {c_\epsilon(\omega) - F(x)}{X_\epsilon(x,\omega)}\, dx$$
where $F$ is an $L^1(]0,1[)$ function (we can add continuity of $F$ on $[0,1]$ if it helps solving the problem) and $c_\epsilon(\omega)$ is defined by
$$c_\epsilon(\omega)\equiv \frac{\int_0^1 \frac{F(y)}{X_\epsilon(y,\omega)} \, dy}{\int_0^1\frac 1 {X_\epsilon(y,\omega)}\, dy}$$
It can be shown that for each $x\in [0,1]$, we have almost surely (we use law of large numbers and the density of the staircase functions in the $L^1$ functions)
$$u_\epsilon(x,\omega) \to u_0(x) \equiv -E[\frac 1 {X_1}](\int_0^x F(y) dy - x \int_0^1 F(y) dy) $$ when $\epsilon \to 0^+$
We would like to find a function $f$ such that
$$\Vert u_\epsilon - u_0\Vert_{L^2(]0,1[ \times \Omega)} \le f(\epsilon) \to 0$$
when $\epsilon \to 0^+$
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