Let $\Omega = [0, 1]$. Let $P$ be the uniform probability law on $\Omega$. Let $X : [0, 1] → \mathbb R$ be a random variable such that $X(t) = t^2$ for all $ t ∈ [0, 1]$. I was asked to calculate $\mathbb E\left[X\,\middle | \left[0,\frac{1}{4}\right]\right]$. I never learnt measure theory and this is what I tried:
$$\mathbb E\left[X\,\middle | \left[0,\frac{1}{4}\right]\right] = \frac{\mathbb E\left(X\cdot \mathbf1_{\left[0,\frac{1}{4}\right]}\right)}{P\left(\left[0,\frac{1}{4}\right]\right)} = \frac1{1/4}\int_0^{\infty}P\left(X\cdot\mathbf 1_{\left[0,\frac{1}{4}\right]} >s\right)ds$$
Then I was told the last step is $$\int_0^{\infty}P\left(X\cdot\mathbf 1_{\left[0,\frac{1}{4}\right]} >s\right)ds=\int_0^{\frac{1}{4}}t^{2}dt$$
I spent a long time but could not figure out how to do the final step. Please help me!
thank you very much!
--------------------------------------------------------------------------------Edit:
I'm probably making unnecessary effort here. My major problem here is how to get the expectation without given pdf. The only way to get expectation that I know is through $$\int xf_X(x)dx$$ or$$\int_0^{\infty}P(X>s)ds$$ for positive random variables. But they both seem not applicable here.