Weakened Chebyshev inequality?

Question

If $f$ is integrable (on some measure space), is it true that:

$$t\nu\{x : f(x) \geq t\}\ \leq \int_{f \geq t}f \, dv\qquad ?$$

$\nu$ is of course the measure, and $t \geq 0$

I'm having trouble proving this, it seems it should be somewhat obvious.

Answer 1

Yes.

Observe that: $$t1_{\{f\geq t\}}\leq f1_{\{f\geq t\}}$$ and take integrals on both sides.