Proposition
Suppose is a complete measure space. Suppose and are s.t. a.e. (almost everywhere). Then if is measurable, is also measurable.
Proof
Let . Then . Consider .
Since a.e., there exists s.t. , and if .
Consider:
Suppose s.t. and .
Since is complete,
and by the measure
Remark:- If is not complete then the proposition may not hold.
Since this is not complete.
, but s.t.
But is complete & so . Consider . Then a.e.
But is not Borel measurable { Borel measurable }
in
Definition (Lebesgue integrable)
Let be a measure space and be measurable. Then is said to be Lebesgue integrable (or just integrable) if
In this case, the Lebesgue integrable of is defined as
Proposition
If are integrable and are disjoint. Then is integrable on , (when well-defined) and are integrable and
i)
ii)
iii) is finite a.e.
iv)
v) If then
vi) If a.e. then
vii) If & then is integrable
Proof
iii) If is not finite a.e. then at least one of the sets
have +ve measure if then
vi) Given a.e. Then , where
Then by ii)