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

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?

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.