If so, then they are both integrable because the integral of a bounded function on a set of finite measure is definitely finite.
12.
A measure space need not be sigma finite, that is, it need not be a countable union of sets of finite measure.
13.
He first defines step functions ( each " step " being a set with finite measure ) and their integrals in the obvious way.
14.
A measure is called " ?-finite " if can be decomposed into a countable union of measurable sets of finite measure.
15.
The class of ?-finite measures has some very convenient properties; ?-finiteness can be compared in this respect to separability of topological spaces.
16.
The ?-finite measure spaces have some very convenient properties; ?-finiteness can be compared in this respect to the Lindel�f property of topological spaces.
17.
Egorov's theorem guarantees that on a finite measure space, a sequence of functions that converges almost everywhere also converges almost uniformly on the same set.
18.
The measures are both decomposable, showing that Tonelli's theorem fails for decomposable measures ( which are slightly more general than ?-finite measures ).
19.
If can be covered by an increasing sequence of open sets that have finite measure, then the space of & ndash; integrable continuous functions is dense in.
20.
The space of measurable functions on a \ sigma-finite measure space ( X, \ mu ) is the canonical example of a commutative von Neumann algebra.
