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wiah

  • one year ago

Prove that if f is measurable, then {x in A|c=f(x)} is measurable for each real number c. Show that the converse is false.

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  1. Raja99
    • one year ago
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    nice pic

  2. Raja99
    • one year ago
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    can u elaborate the question!

  3. wiah
    • one year ago
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    if f(x) is measurable on a set A, then A(f(x)=c) is measurable for each c.how to prove this?

  4. Raja99
    • one year ago
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    take a example and prove

  5. Raja99
    • one year ago
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    u need have some functions for this

  6. wiah
    • one year ago
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    do u know any link regarding measurable functions?i need to know the proof of all the properties of measurable functions.

  7. Raja99
    • one year ago
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    nope

  8. Edutopia
    • one year ago
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    http://en.wikipedia.org/wiki/Measurable_function Inverse image "Preimage" redirects here. For the cryptographic attack on hash functions, see preimage attack. Let f be a function from X to Y. The preimage or inverse image of a set B ⊆ Y under f is the subset of X defined by The inverse image of a singleton, denoted by f −1[{y}] or by f −1[y], is also called the fiber over y or the level set of y. The set of all the fibers over the elements of Y is a family of sets indexed by Y. Again, if there is no risk of confusion, we may denote f −1[B] by f −1(B), and think of f −1 as a function from the power set of Y to the power set of X. The notation f −1 should not be confused with that for inverse function. The two coincide only if f is a bijection.

  9. Edutopia
    • one year ago
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    http://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch3.pdf

  10. Edutopia
    • one year ago
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    http://mathforum.org/library/drmath/view/66541.html

  11. watchmath
    • one year ago
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    If f is measurable, then for any c, A_c={x | f<c} is measurable and so is the complemant \(A_c^C\). It follows that {x | f(x) =c} =A\ (A_c U A_c^C) is also measurable.

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