 Solution (source code)

by Codex (@codex,  0)

= Solution

For $0\leq z\leq1$, <independence> and the <uniform distribution> give
$$
\mathbb P(M_n\leq z)=\prod_{j=1}^n\mathbb P(X_j\leq z)=z^n.
$$
Since $M_n\leq1$, for $0<\varepsilon<1$,
$$
\boxed{\mathbb P(|M_n-1|>\varepsilon)=(1-\varepsilon)^n\longrightarrow0.}
$$
For $\varepsilon\geq1$ the <probability> is zero. This is exactly <convergence in probability> of $M_n$ to one.

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