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proof of Egorov's theorem
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(Proof)
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Let
Since almost everywhere, there is a set with such that, given
and , there is
such that implies
. This can be expressed by
or, in other words,
Since
is a decreasing nested sequence of sets, each of which has finite measure, and such that its intersection has measure 0, by continuity from above we know that
Therefore, for each
, we can choose such that
Let
Then
We claim that uniformly on
. In fact, given
, choose such that
. If
, we have
which in particular implies that, if ,
; that is,
. Hence, for each
there is (which is given by above) such that implies
for each
, as required. This completes the proof.
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"proof of Egorov's theorem" is owned by Koro.
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(view preamble)
Cross-references: intersection, measure, finite, sequence, decreasing, implies, almost everywhere
This is version 4 of proof of Egorov's theorem, born on 2003-07-27, modified 2006-07-27.
Object id is 4518, canonical name is ProofOfEgorovsTheorem.
Accessed 4663 times total.
Classification:
| AMS MSC: | 28A20 (Measure and integration :: Classical measure theory :: Measurable and nonmeasurable functions, sequences of measurable functions, modes of convergence) |
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Pending Errata and Addenda
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