Markov's inequality states:
Let $X$ be a non-negative random variable and suppose that $\mathbb{E}(X)$ exists. For any $t > 0$:
$$\mathbb{P}(X > t) \leq \frac{\mathbb{E}(X)}{t}$$
My text contains the following proof:
Since $X > 0$,
$$ \begin{align} \mathbb{E}(X) &= \int_0^{\infty} xf_X(x)dx \\ &= \int_0^{t} xf_X(x)dx + \int_t^{\infty} xf_X(x)dx \\ &\geq \int_t^{\infty} xf_X(x)dx \\ &\geq t\int_t^{\infty} f_X(x)dx \\ &= t\mathbb{P}(X > t) \end{align} $$
My concern is the step where we remove $x$ from the integral. I think the assumption is that since we know $x$ is non-negative, that removing multiplication by $x$ can only make things smaller. However, for $0 < x < 1$ removing multiplication by $x$ should actually make them bigger. If $f_X$ only has density in that range, then I don't think you can say that $\int_t^{\infty} xf_X(x)dx \geq \int_t^{\infty} f_X(x)dx$. Unless somehow multiplying by $t$ at the same time gets rid of this problem?
