Integrals of indicator functions question

Question

I have a result

$\int_X \int_Y \mathbb{1}[h(x,y) < \mu]dP(y)dP(x) < a$

and I am trying to resolve the integral

$\int_X \int_Y \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}] \mathbb{1}[h(x,y) < \mu] dP(y)dP(x)$

From some previous work I can use Markov's inequality, $P[|f(x) - g(x)| > \frac{\mu}{2}] < \frac{2 \sqrt{\epsilon}}{\mu}$

I only have a limited understanding of the relationship between integrals, indicator functions, and am trying to understand how to use the two results I have to get an upper bound on the integral in question. I know (think) that $\int_X \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}]dP(x) = P[|f(x) - g(x)| > \frac{\mu}{2}$, and that the function of just $x$ should be constant w.r.t. the $Y$ integral, but how can I resolve this when one of my applicable results only goes halfway through the double integral and the other goes all the way through?

Answer 1

It depends what kind of an upper bound you're looking for.

Note that at least we have $\mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}]\leq 1$, so \begin{align*} &\int_X \int_Y \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}] \mathbb{1}[h(x,y) < \mu] dP(y)dP(x) \\ &\leq \int_X \int_Y \mathbb{1}[h(x,y) < \mu] dP(y)dP(x)<a. \end{align*} If you want something sharper, then use Cauchy-Schwarz $(1)$, Jensen's (note that $x\mapsto x^{2}$ is convex) $(2)$, and then your given upper bounds that $(3)$, to get \begin{align*} &\int_X \int_Y \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}] \mathbb{1}[h(x,y) < \mu] dP(y)dP(x) \\ &=\int_X \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}]\int_Y \mathbb{1}[h(x,y) < \mu] dP(y)dP(x) \\ &\overset{(1)}{\leq} \Big(\int_X \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}]^{2}\,dP(x)\Big)^{\frac{1}{2}}\Big(\Big(\int_Y \mathbb{1}[h(x,y) < \mu] dP(y)\Big)^{2}dP(x)\Big)^{\frac{1}{2}} \\ &\overset{(2)}{\leq} \Big(\int_X \mathbb{1}[|f(x) - g(x)| > \frac{\mu}{2}]^{2}\,dP(x)\Big)^{\frac{1}{2}}\Big(\int_Y \mathbb{1}[h(x,y) < \mu] ^{2}dP(y)dP(x)\Big)^{\frac{1}{2}} \\ &=P[|f(x) - g(x)| > \frac{\mu}{2}]^{\frac{1}{2}}\Big(\int_Y \mathbb{1}[h(x,y) < \mu] dP(y)dP(x)\Big)^{\frac{1}{2}}\\ &\overset{(3)}{\leq} \Big(\frac{2\sqrt{\varepsilon}}{\mu}\Big)^{\frac{1}{2}}\sqrt{a}. \end{align*}