finite measure sentence in Hindi
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- For finite measures and, the idea is to consider functions with.
- We wish to obtain a generally finite measure as the bin size goes to zero.
- More generally the spaces with an atomless, finite measure and are not locally convex.
- This property is false without the assumption that at least one of the has finite measure.
- It turns out that | & mu; | is a non-negative finite measure.
- If is a finite measure on, the function admits for the convergence in measure the following fundamental system of neighborhoods
- The fact that the remaining part of is singular with respect to follows from a technical fact about finite measures.
- Conversely, any homogeneous system of imprimitivity is of this form, for some measure ?-finite measure ?.
- For instance, it is used in proving the uniqueness claim of the Carath�odory extension theorem for ?-finite measures.
- Once the result is established for finite measures, extending to-finite, signed, and complex measures can be done naturally.
- If so, then they are both integrable because the integral of a bounded function on a set of finite measure is definitely finite.
- A measure space need not be sigma finite, that is, it need not be a countable union of sets of finite measure.
- He first defines step functions ( each " step " being a set with finite measure ) and their integrals in the obvious way.
- A measure is called " ?-finite " if can be decomposed into a countable union of measurable sets of finite measure.
- The class of ?-finite measures has some very convenient properties; ?-finiteness can be compared in this respect to separability of topological spaces.
- The ?-finite measure spaces have some very convenient properties; ?-finiteness can be compared in this respect to the Lindel�f property of topological spaces.
- 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.
- The measures are both decomposable, showing that Tonelli's theorem fails for decomposable measures ( which are slightly more general than ?-finite measures ).
- 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.
- The space of measurable functions on a \ sigma-finite measure space ( X, \ mu ) is the canonical example of a commutative von Neumann algebra.
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