Any semi closed interval is contained in the borel $\sigma$-field on $\mathbb{R}$

35 Views Asked by At

I would like to test my intuition of the borel $\sigma$-field on $\mathbb{R}$ $(B(\mathbb{R}))$ by proving the above statement. My proof is as follows: I look at the case for (a,b]

clearly $(a,\infty) \in B(\mathbb{R})$ , thus

$(-\infty,a] \in B(\mathbb{R})$.

similarly, $(b,\infty) \in B(\mathbb{R})$

we put it together and get that $$(-\infty,a] \bigcup (b,\infty) \in B(\mathbb{R})$$ by definition. Thus its complement is also in $B(\mathbb{R})$.

Is the reasoning correct? Any help is appreciated.

1

There are 1 best solutions below

0
On BEST ANSWER

Yes it is correct. By definition unions and compliments exists in $\sigma$-algebra, and if we combine them by De-Morgan's theorem Intersections also make into $\sigma$-algebra.