sub interval on R generates a borel sigma algebra proof

Question

If anyone could help me with proof of the set $\lbrace (−\infty ,x] : x \in \mathbb{R} \rbrace$ generates a Borel $\sigma$-algebra ?

Answer 1

HINT: It generates sets $(-\infty,x)$, $(y,\infty)$, and so all open intervals.

Answer 2

First you have a set that It is

I = { (−∞,x] c ℝ : x ∈ ℝ}

Then you create a new set to obtain the emptyset

I' ={]a,b] c ℝ: -∞ <= a <= b <= +∞} here is te emptyset given that this possible interval belongs to I' and it's an emptyset ]a,a] = 0

then we notice that the complementar ]a,b] is not in I' Note ]a,b]c =]-∞,a]U]b,+∞]. So we create a Borel algebra to put all the unions

A(ℝ)= {U In c R: I1,...,In}