= Solution
For a continuous strictly increasing <cumulative distribution function> $F$ and a <uniform distribution> $U$ on $(0,1)$, monotonicity gives
$$
\Pr(F^{-1}(U)\le y)=\Pr(U\le F(y))=F(y).
$$
Thus \b[each transformed draw has distribution function $F$]. This is <inverse transform sampling>. The conclusion extends to discontinuous or non-strictly increasing $F$ by its <quantile function> $Q(u)=\inf\{x:F(x)\ge u\}$: right continuity gives $Q(u)\le y$ exactly when $u\le F(y)$. Independent uniforms produce independent transformed draws, but marginal uniformity alone does not establish <independence>.
If “uniformly distributed sequence” instead means deterministic equidistribution, the empirical fraction of $\eta_i\le y$ is the empirical fraction of $\xi_i\le F(y)$ and tends to $F(y)$. That is an empirical distribution statement, not a claim that the deterministic sequence consists of independent <random variables>.
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