While studying measure theory, I came across Markov's inequality, which states that given a measure space $(X,\mathcal{S},\mu)$ and a function $h \in \mathcal{L}^1(\mu)$, we have
$$\mu \left( \left\{x \in X \colon |h(x)| \geq c \right\} \right) \leq \frac{1}{c} \|h\|_1$$
for all $c>0$. Under what conditions does equality hold?
It seems that one case would be when $\|h\|_1 = 0$, but I do not see any reason for equality to hold for $\|h\|_1 > 0$. Any help would be greatly appreciated. Thank you.
Best Answer
$$ \frac{\Vert h\Vert_1}{c}-\mu(\{\vert h\vert\ge c\})=\int_X\left(\frac{\vert h(x)\vert}{c}-1_{\{\vert h\vert\ge c\}}\right)\,\mu(dx). $$
Markov's inequality is an equality iff the integral above is zero. Yet it is clear that the integrand is nonnegative, hence the integral is zero iff the integrand is $\mu$-a.e. zero. In other words, you have equality iff $\mu(dx)$-a.e., $\vert h(x)\vert=c1_{\{\vert h\vert\ge c\}}$.
One can readily show that the latter condition is equivalent to $\vert h(x)\vert\in\{0,c\}$, $\mu(dx)$-a.e.