§ 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(XA)P(X \geq A). Since we're in measure land, we have no way to directly access P()P(\cdot). The best we can do is to integreate the constant function 11, since the probability is "hidden inside" the measure. This makes us compute:
P(XA)={XA}1dμ P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu
Hm, how to proceed? We can only attempt to replace the 11 with the XX to get some non-trivial bound on XX. But we know that XAX \geq A. so we should perhaps first introduce the AA:
P(XA)={XA}1dμ=1/A{XA}Adμ 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 XX:
P(XA)={XA}1dμ=1/A{XA}Adμ<1/A{XA}Xdμ=1/AE[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(XA)E[X]/A P(X \geq A) \leq \mathbb{E}[X]/A
So we are "smearing" the indicator 11 over the domain {XA}\{X \geq A\} and attempting to get a bound.