If $w \ge 0$ and $\mu$ is a finite measure on $X$ with $\mu(w = 0) \ge \mu(X)/2$, is $\mu(w > t) \le 2 \mu(|w - c| > t/2)$?

Question

Suppose we have a finite measure space $(X, \mu)$ and a measurable function $w \ge 0$ on $X$ satisfying $\mu(w = 0) \ge \mu(X)/2$, is it true that for any real $c$ we have $\mu(w > t) \le 2 \mu(|w - c| > t/2)$ whenever $t > 0$?

If $c > t/2$ then clearly $\{ w = 0 \} \subset \{c - w > t/2\}$, but $c - w(x) > t/2 > 0$ implies $|w(x) - c| = c -w(x) > t/2$, so we have $$ \mu(X)/2 \le \mu(w = 0) \le \mu(|w -c| > t/2). $$ On the other hand, $\{ w > t\} \subset \{ w \neq 0\}$ so $$\mu(w > t) \le \mu(X) - \mu(w = 0) \le \mu(X)/2 \le \mu(|w - c| > t/2), $$ which proves the claim when $c > t/2$.

When $c \le t/2$, we have $w > t \ge c + t/2$ so $\{ w > t\} \subset \{w - c > t/2\} \subset \{ |w - c| > t/2\}$, from which we again see that $\mu(w > t) \le \mu(|w - c| > t/2)$.

However, in Petersen's Riemannian Geometry (in his proof of Theorem 7.1.13), and in Lemma 2.2 of this paper, the inequality given is $$ \mu(w > t) \le 2 \mu(|w -c| > t/2), $$ this extra factor of $2$ leads me to believe that I must have done something wrong, but the argument is so simple I cannot find any problems with it. Am I missing something obvious?

Answer 1

What you have written seems correct to me.

I suspect that what has happened here is that the inequality that includes the $2$ was sufficient for the purposes of the authors of that paper (which was Petersen's source for the proof) and that inequality is even more straightforward to obtain; you can remove the sentence in your question starting "on the other hand" and simply write $$\mu(w > t) \le \mu(X) \le 2 \mu(|w-c|>t/2)$$ in this case by the line above.