Question

Show that every continuous function is measurable.

Answer 1

If a function f:Rm→Rn is continuous, then it is measurable. Proof. ... The set of measurable functions is strictly larger than the set of continuous functions. For example, the indicator function IA:R→R, defined as IA(x)=1 if x∈A and 0 otherwise, is not continuous (assuming A≠Ω and A≠∅) but is measurable if A∈B(R).

Answer 2

Answer:

A function f: RmRn is measurable if it is continuous. There is proof... The set of measurable functions exceeds the set of continuous functions by a factor of two. For instance, the indicator function IA: RR, which is defined as IA(x)=1 if xA and 0 otherwise, is not continuous (assuming A and A) but is measurable if AB (R)

Step-by-step explanation:

All continuous functions are measurable when using the Lebesgue measure, or more broadly any Borel measure. In fact, every function that can be specified may be measured. Addition and multiplication close measurable functions, while composition does not.

How can you show that a continuous function can be measured:

A function f: RmRn is measurable if it is continuous. Proof. The collection of all open sets generates the -algebra B(Rn).

Is there a continuous measurable function practically everywhere?

Lusin's theorem (or Luzin's theorem, named after Nikolai Luzin) or Lusin's criteria is a mathematical theorem that asserts that an almost everywhere finite function is measurable if and only if it is a continuous function over nearly all of its domain.