Proof of Markov's inequality using alternate form of expectation For nonnegative random variables $X$, there is an alternate expression for the expectation: $$E[X] = \int_0^\infty P(X \ge t) \mathop{dt}.$$ I am familiar with proofs of Markov's inequality $$P(X \ge t) \le \frac 1t E[X]$$ that use the usual definition of expectation $E[X] := \int_0^\infty t \cdot f(t)\mathop{dt}$ where $f$ is the pdf of $X$. However, is there an easy way to prove it using the special form of the expectation above (and without reverting to the usual definition)?

$$ \mathbb{E}\left[X \right] = \int_0^{\infty} P\left(X\geq u \right) du \geq \int_0^{t} P\left(X\geq u \right) du \geq\int_0^{t} P\left(X\geq t \right) du = tP\left(X\geq t \right).$$