Expected value of non-negative bounded random variable

Question

I have the random variable $Y$ and it is such that $0\leq Y\leq 1$ I want to bound its expected value, given that I have a high probability bound: $P(Y\geq x) \leq g(x)$, can I say that: $$ E[Y] = \int_0^1P(Y\geq x) dx \leq \int_0^1 g(x) dx\, ?$$ I tried showing it but I am not really strong in measure theory and Fubini theorem: \begin{align}\int_0^1P(Y\geq x) dx &= \int_0^1 \left(\int_y^\infty f_Y(z)dz\right)dy \\ &= \int_0^1 \left(\int_0^z f_Y(z)dy\right)dz \\ &= \int_0^1 z\, f_Y(z)dz = E[Y] \end{align} Is it correct?

Answer 1

Replacing $f_Y(z)dz$ by $dF_Y(z)$ would make your proof valid for any $Y$, whether or not $f_Y$ exists. [To be precise you will have to write $\int_y^{\infty}d_F(z)$ as $\int_{[y,\infty)} dF_Y(z)$].