Question

If f:R -> R is lebesgue measurable and g:R-> R is continuous. then g(f(x)):R->R is lebesgue measurable.

Answer 1

@eliassaab it seems to me that this should say f(g(x)) is lebesgue measurable. is the way I asked in the question true?

Answer 2

again, if you are sick of me tagging you, let me know. But I figure I am keeping you sharp on your measure theory:)

Answer 3

Thank you for keeping me sharp on something that I did all my life

Answer 4

lol

Answer 5

Do you know how to do it?

Answer 6

nope. I thought I did, but i can do f(g(x)), but not quite sure about g(f(x))

Answer 7

so I need to show that the pre image of any open set is in M. so the preimage of f of an open set is lebesgue measurable but im not sure how to show that the preimage of g of a lebesgue measurable set is lebesgue measurable. but all I have is that g is continuous.

Answer 8

Let M be an open set
\[

(g f)^{-1} (M) = f^{-1}( g^{-1} (M))
\]
Since \[ g^{-1} (M) \] is open then \[
f^{-1}( g^{-1} (M))
\]
is measurable and hence fg is measurable. The other way is not true in general, see
http://math.stackexchange.com/questions/283443/is-composition-of-measurable-functions-measurable

Answer 9

so if g is continuous and f is Lebesgue measurable then its not true (in general) that \(g\cdot f\) is Lebesgue measurable?