1) { }=Q , E={x X : g(x)<f(x)}

Show that E= [{x X : g(x)< } {x X : f(x)> }]

2) {x X : g(x)+f(x)<a} = {x X : g(x)<a-f(x)}.

If f(x) is a countable function show that a-f(x) is countable function too.

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- May 12th 2009, 10:46 AMSENTINEL42 exercises
1) { }=Q , E={x X : g(x)<f(x)}

Show that E= [{x X : g(x)< } {x X : f(x)> }]

2) {x X : g(x)+f(x)<a} = {x X : g(x)<a-f(x)}.

If f(x) is a countable function show that a-f(x) is countable function too. - May 12th 2009, 11:03 AMPlato
- May 12th 2009, 11:40 AMHallsofIvy
Also, what do you mean by "countable function"? I know what "countable sets" are but not "countable function". It may be a translation problem. Do you mean "measurable function".

- May 12th 2009, 12:21 PMSENTINEL4
@Plato

Under the is not n.

@HallsofIvy

It means measurable not countable. - May 12th 2009, 12:37 PMPlato