# If $f$ is non-increasing, why is $\mathbb{P}(\ f(X+\epsilon) \leq f(X_n) \leq f(X-\epsilon)) \leq \mathbb{P}(X-\epsilon < X_n < X+\epsilon)$?

If $f$ is a non-increasing non-random function and $X_n$ is a sequence of random variables, why do we have that:

$$\mathbb{P}\left[\ f(X+\epsilon) \leq f(X_n) \leq f(X-\epsilon)\right] \leq \mathbb{P}\left[X-\epsilon < X_n < X+\epsilon\right]$$ ?

I understand that the conventional way to to use events and subsets to show that:

$$\{\omega \in \Omega: f(X+\epsilon) \leq f(X_n(\omega)) \leq f(X-\epsilon)\}\subseteq \{\omega \in \Omega: X-\epsilon < X_n(\omega) < X+\epsilon\}$$

In other words, an event $\omega$ in the first set is an event in the second set. HOWEVER, it seems that if we let $f$ be a constant function, then there may exist an $\omega' \in \Omega$ where we can have $X_n(\omega') = X+\epsilon$, and so:

$$\omega' \in \{\omega \in \Omega: f(X+\epsilon) \leq f(X_n(\omega)) \leq f(X-\epsilon)\}$$

BUT

$$\omega' \notin \{\omega \in \Omega: X-\epsilon < X_n(\omega) < X+\epsilon\}$$

?

Is there something going on here I am missing? Thanks!

You might have that backwards. You can "do algebra" within the parentheses of $\mathbb{P}$. Also $A \subset B$ implies $\mathbb{P}(A) \le \mathbb{P}(B)$. \begin{align*} \mathbb{P}\left[X-\epsilon < X_n < X+\epsilon\right] &= \mathbb{P}[(f(X-\epsilon) > f(X_n) > f(X + \epsilon)] & \text{ (f non-increasing)}\\ &\le \mathbb{P}[(f(X-\epsilon) \ge f(X_n) \ge f(X + \epsilon)]. & (\text{subsets}) \end{align*} The last line works because those two sets are subsets. If one thing is bigger (for a particular $\omega$) than the other strictly, then in particular it's bigger than or equal to it.
• So basically the last line's set is larger due to the equality and so in a sense we have more "cases" for $\omega$? Commented May 1, 2017 at 5:46