# Is this proof of Markov's inequality correct?

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?

Note that the range of integration is $$x = t$$ to $$x=\infty$$. On that range, you have $$x \geq t$$, so you can lower bound $$x$$ by $$t$$. Since the density is always nonnegative, $$f_X(x) \geq 0$$ for any $$x$$, you can write $$xf_X(x) \geq tf_X(x)$$ whenever $$x \geq t$$.
To elaborate, the calculation is $$\int_t^\infty xf_X(x) dx \geq \int_t^\infty tf_X(x)dx = t \int_t^\infty f_X(x)dx.$$