If $\{x\in X: f(x)>a\}$ is measurable then $\{x\in X: f(x)\geq a\}$ is measurable If $\\{x\in X: f(x)>a\\}$ is measurable then $\\{x\in X: f(x)\geq a\\}$ is measurable We know that $\\{x\in X: f(x)>a-\frac{1}{n}\\}$ is measurable, so it is in $S$ which is a sigma algebra, We do $\\{x\in X: f(x)\geq a\\}=\cap_{n=1}^{\infty}\\{x\in X: f(x)>a-\frac{1}{n}\\}$? in particule why does this intersection includes $f(x)=a$?

The direction \begin{align} \\{x\in X:f(x)\geq a\\}\subseteq\bigcap_{n=1}^{\infty}\left\\{x\in X:f(x)>a-\frac{1}{n}\right\\} \end{align} should be clear. Indeed, let $x$ be an element of the L.H.S., so that $f(x)\geq a$. Since $a>a-\frac{1}{n}$, this implies that $f(x)>a-\frac{1}{n}$ for _every_ $n\in\mathbb{N}$, hence $x$ is in the R.H.S. as well.

For the other direction, it would also be good to prove in the contrapositive form. Suppose $x_0$ is _not_ an element of the L.H.S., so that $f(x_0)0$, there exists a sufficiently big $N\in\mathbb{N}$ such that $\delta>\frac{1}{N}$. Then it follows that \begin{align} f(x_0)&otin\left\\{x\in X:f(x)>a-\frac{1}{N}\right\\}$, hence it cannot be in the intersection, which is the R.H.S..