Does |1/x| have an essential supremum on (-1, 1)?

[futurebird]

I want to say that f(x) = |1/x| is in L-infinity(E) when m(E)<infinity because the function has and esssup on any measurable set, E. Even if E = (-1, 1) f(0) is not a problem since it is only one point...

But wait... what *is* the esssup for this function on (-1, 1)? I think it might not have one. This is why I'm confused.

:(

 

[LCKurtz]

You don't seem too confused to me. It doesn't have an esssup on (-1,1).

 

[futurebird]

Ok. So then it's not in L-infinity when x=0 is in E.

(slowly this starts to make more sense...)

 

[LCKurtz]

Not exactly. E could be (-oo, -1) union [0] union (1,oo) and it would be in L^oo.

 

[futurebird]

oh good point.

 

[LCKurtz]

You could have f(x) = sin(x) except on the rationals where it is 1/x and anything at 0. This would have an esssup of 1 because the set of x where |f(x)| > 1 has measure 0.

 

[futurebird]

Thanks, that's a good example to think about.