Aim
If , then
Define
is a metric space with metric
Theorem
is complete with respect to metric .
Note
To show completeness of a metric space , it is enough to check convergence of Cauchy sequences satisfying
Proof
Start with a Cauchy sequence in such that
Define, for ,
and
By Minkowski inequality,
Hence a.e. and by MCT,
So . Therefore there exists with such that for all .
Define
Since a.e. and , it follows that .
Also, for each ,
Define
Then and for all (hence a.e.).
Now for ,
So
Hence
Fix and apply Fatou’s lemma:
Since a.e., we get
Therefore
so in .
Hence is complete.
If is Cauchy in , then there exists a subsequence such that
By the above proof, in for some .
In fact, we also proved pointwise a.e.
If in , then there exists a subsequence such that
Example (No pointwise convergence in general)
Take , (Lebesgue measure), and define indicator functions by scanning dyadic-type subintervals, e.g.
For , one has
hence in .
But if , then for infinitely many .
So pointwise.
Essentially Bounded Functions
A measurable is essentially bounded if there exists such that
Define