# Measurable set

#### Fermat

##### Active member
I am trying to show that an open set in [0,1] is measurable, given that [0,x] is measurable set for each x in [0,1]. So I need to show (a,b) is measurable. Using the fact that measurable sets form a sigma algebra, I have managed to show that (a,b] is measurable. So (a,b+t] is measurable for any t>0. letting t ->0, can I then conclude that (a,b) is measurable? It seems a bit easy that I can just relax closed intervals to open intervals in this way.

#### HallsofIvy

##### Well-known member
MHB Math Helper
You are given, then, that (0, a) and (0, b) are measurable and you can write (a,b)= (0, b)\ (0, a).

#### Opalg

##### MHB Oldtimer
Staff member
I am trying to show that an open set in [0,1] is measurable, given that [0,x] is measurable set for each x in [0,1]. So I need to show (a,b) is measurable. Using the fact that measurable sets form a sigma algebra, I have managed to show that (a,b] is measurable. So (a,b+t] is measurable for any t>0. letting t ->0, can I then conclude that (a,b) is measurable?
Not quite, there are two things wrong here. First, in a sigma-algebra, you must use countable unions and intersections. So instead of $t$ you should have $1/n$. Then you can take the intersection of the sets $\bigl(a,b+\frac1n\bigr]$. That leads us to the second thing that is wrong, namely the fact that $$\displaystyle \bigcap_n \bigl(a,b+\tfrac1n\bigr] = (a,b]$$: you still have an interval that is closed at the upper end!

What you need to do is to take the union of the sets $\bigl(a,b-\frac1n\bigr]$, which is $(a,b)$.