A basic question from measure theory I'm very new in measure theory and I have the following question:
Let $(X, \Sigma, \mu$) be a probability measure, that is, $\mu (X)=1$. Two measurable, real valued functions  $f$ and $g$ on $X$ are called independent if for all $ \alpha , \beta \in \mathbb{R}$, $\mu( \{ f<\alpha \})\cap \mu( \{ g<\beta \})=\mu( \{ f<\alpha \})\mu( \{ g<\beta \})$
Notation: $\{ f<\alpha \}=f^{-1}(-\infty,\alpha )$
Show that if $f$ and $g$ are independent, then for all intervals $I$ in $\mathbb{R}$ and $\beta \in \mathbb{R}$, we have $\mu(f^{-1}(I)\cap  \{g<\beta \})=\mu(f^{-1}(I))\mu (\{g<\beta \})$
First I try to show for any semi-open intervals $I=[\alpha, \gamma)=(-\infty,\gamma)\cap[\alpha,\infty)$. So $f^{-1}(I)= \{ f<\alpha \}\cap \{ f\geq\alpha \}$. But here I cannot think for the rest. Thanks!

For a bounded measure(such as a probability measure), we have $$A_n \uparrow A \implies P(A_n) \uparrow P(A)$$ and $$A_n \downarrow A \implies P(A_n) \downarrow P(A)$$ So you can generalize your independence condition to non-strict inequalities such as
$$ \mu( \{ f\leq\alpha \}\cap \{ g<\beta \})=\mu( \{ f\leq \alpha \})\mu( \{ g<\beta \})$$
and other similar versions.
Then for example if $I = (a,b) = (-\infty, b) - (-\infty,a)$, we have
\begin{align}
\mu(f^{-1}(I)\cap \{g<\beta \})&= \mu(f^{-1}(-\infty, b)\cap \{g<\beta \}) - \mu(f^{-1}(-\infty, a)\cap \{g<\beta \})\\
& = \mu( \{ f < b \})  \mu( \{ g < \beta \}) - \mu( \{ f < a \})  \mu( \{ g < \beta \}) \\
& = (\mu( \{ f < b \}) - \mu( \{ f < a \})) \mu( \{ g < \beta \}) \\
& = \mu(f^{-1}(I)) \mu( \{ g < \beta \})
\end{align}
Similarly, we can get the result for closed and semiclcosed interval
