Consider a random variable $X$ with $E[X]< \infty$ and probability density $f_X$, I was wondering: given any positive number $\epsilon >0,$ do we always have $$ \int_{-1/\epsilon}^\infty x f_X(x) dx \le E[X]? $$
Here is my thinking: Given any $\epsilon >0, $ $$ \begin{align} \int_{ - 1/\varepsilon }^\infty x {f_X}(x)dx &= \int_{ - \infty }^\infty {{1_{\{ x \ge - 1/\varepsilon \}}}x} {f_X}(x)dx \le ? \le E[X] \end{align}$$ It is tempting to say yes but my $X$ can take negative values; e.g., say $X$ is normal distributed, in this case I kind of stuck to argue further.
For any $\epsilon > 0$,
$$ \begin{align*} E[X] & = \int_{-\infty}^{+\infty} xf_X(x)dx \\ & = \int_{-\infty}^{-\epsilon} xf_X(x)dx + \int_{-\epsilon}^{+\infty} xf_X(x)dx \\ & \leq \int_{-\epsilon}^{+\infty} xf_X(x)dx \end{align*}$$
as the integrand $xf_X(x)$ is negative on $(-\infty, -\epsilon)$ as pointed out by others.