Converge in probability

Question

I'm trying to solve the following Probability problem:

Consider $ X(\omega) : [0,1] \to [0,1] $, where $f(x) = 2x \mathbb{I}_{[0,1/2]}(x) + 0.75 \mathbb{I}_{\{1\}}(x) $ and $X(\omega) = \omega = x$. Consider $X_n = X^n$. Prove that $X_n$ does not converge in probability to $0$.

This is what I've got so far:

$\lim_{n \to \infty} P \left( \{ \omega \in \Omega : |X_n(\omega) - 0| > \varepsilon \} \right) = \lim_{n \to \infty} P \left( \{ \omega \in \Omega : X_n(\omega) > \varepsilon \} \right) = \lim_{n \to \infty} P \left( X^n > \varepsilon \right) = 1 - \lim_{n \to \infty} P \left( X \leq \varepsilon^{1/n} \right) $.

I divided this in three cases:

1. $\varepsilon > 1$, then $P(X \leq \varepsilon^{1/n}) = 1 $

2. $\varepsilon < 1$, then $P(X \leq \varepsilon^{1/n}) = \int_{0}^{\varepsilon^{1/n}} 2x \; dx = \varepsilon^{2/n}$

3. $\varepsilon = 1$ (I think that the problem might be here!) $P(X \leq 1) = 1$ because I'm integrantig all over the support of X.

With this, I conclude:

$\lim_{n \to \infty} P \left( \{ \omega \in \Omega : |X_n(\omega) - 0| > \varepsilon \} \right) = 1 - \lim_{n \to \infty} P \left( X \leq \varepsilon^{1/n} \right) = \begin{cases} 1 - 1 = 0 & \text{ if } \varepsilon > 1 \\ 1 - \lim_{n \to \infty} \varepsilon^{2/n} = 1 - 1 = 0 & \text{ if } \varepsilon < 1 \\ 1 - 1 = 0 & \text{ if } \varepsilon = 1 \end{cases} $

So I looks like it does converge in probability to $0$, but that's exactly the opposity to what I was asked to show. I appreciate your comments.

Thank you!

Answer 1

Your issue is actually in the second case. The limits of your integral from $0$ to $\varepsilon^{1/n}$ ignores the fact that there is an indicator $\mathbb I_{[0, 1/2]}(x)$ multiplied to the integrand $2x$. This will change the nature of the integration for large enough $n$ because $\varepsilon^{1/n}$ eventually surpasses $1/2$:

If $0 < \varepsilon < 1$, then there exists an $N \in \mathbb N$ such that for all integers $n \geq N$ we have $\varepsilon^{1/n} > 1/2$. Thus for $n \geq N$ $$ P[X_n \leq \varepsilon] = P[X \leq \varepsilon^{1/n}] = \int_{0}^{\varepsilon^{1/n}} 2x \mathbb I_{[0, 1/2]}(x) \ dx = \int_{0}^{1/2} 2x \ dx = 0.25 $$ and this shows that $X_n \not\to_p 0$ as $P[|X_n| > \varepsilon] = 1 -0.25 = 0.75$ constantly for all $n \geq N$.