## § Markov and chebyshev from a measure theoretic lens

I've been idly watching Probability and Stochastics for finance: NPTEL , and I came across this nice way to think about the markov and chebyshev inequality. I wonder whether Chernoff bounds also fall to this viewpoint.

#### § Markov's inequality

In markov's inequality, we want to bound $P(X \geq A)$. Since we're in measure land, we have no way to directly access $P(\cdot)$. The best we can do is to integreate the constant function $1$, since the probability is "hidden inside" the measure. This makes us compute:
$P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu$
Hm, how to proceed? We can only attempt to replace the $1$ with the $X$ to get some non-trivial bound on $X$. But we know that $X \geq A$. so we should perhaps first introduce the $A$:
$P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = 1/A \int_{\{X \geq A\}} A d \mu$
Now we are naturally led to see that this is always less than $X$:
\begin{aligned} &P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = \\ & 1/A \int{\{X \geq A\}} A d \mu < 1/A \int_{\{X \geq A\}} X d \mu = 1/A \mathbb{E}[X] \end{aligned}
This completes marov's inequality:
$P(X \geq A) \leq \mathbb{E}[X]/A$
So we are "smearing" the indicator $1$ over the domain $\{X \geq A\}$ and attempting to get a bound.