Show that if f is a measurable real-valued function and g is a continuous function defined on $\displaystyle (-\infty ,\infty )$, then $\displaystyle g\circ f$ is measurable.

Please help! :'(

- Sep 30th 2009, 03:01 PMDark Sunthe composition of a measurable function with a continuous function is measurable
Show that if f is a measurable real-valued function and g is a continuous function defined on $\displaystyle (-\infty ,\infty )$, then $\displaystyle g\circ f$ is measurable.

Please help! :'( - Sep 30th 2009, 06:18 PMDark Sun
This is what I have developed in a moment of clarity; Please tell me if I am right or wrong:

Certainly every continuous function is measurable, since every interval is measurable (P. 67). Therefore, $\displaystyle g$ is measurable.

Since $\displaystyle f$ is a real-valued function, restrict $\displaystyle g$ from $\displaystyle (-\infty ,\infty )$ to the co-domain of $\displaystyle f$, which is a subset of $\displaystyle (-\infty ,\infty )$. Since $\displaystyle f$ is measurable, the domain of $\displaystyle g$ restricted to the co-domain of $\displaystyle f$ has a measurable domain, that is, the set of all $\displaystyle f(x)$.

Also, the set of all elements not in the domain of $\displaystyle g|_{f(x)}$ are $\displaystyle (-\infty ,\infty )\sim f(x)$, which must neccessarily be measurable.

Therefore, since $\displaystyle g$ is measurable and $\displaystyle [(-\infty ,\infty )\sim f(x)]\cup f(x)=(-\infty ,\infty )$, the domain of $\displaystyle g$, and both $\displaystyle f(x)$ and $\displaystyle (-\infty ,\infty )\sim f(x)$ are measurable, then by Problem 21.a., the function $\displaystyle g$ restricted to the co-domain of $\displaystyle f$ must be measurable. Then $\displaystyle g(f(x))=g\circ f(x)$ is measurable since for all $\displaystyle x$, an element in the co-domain of $\displaystyle f$ is defined, and $\displaystyle g$ restricted to these elements is measurable. - Oct 1st 2009, 07:07 PMputnam120
Look at the inverse of $\displaystyle g\circ f$. So $\displaystyle f^{-1}\circ g^{-1}(U)$ is measurable (U measurable) since, g is continuous and thus measurable. (pull backs of open sets by $\displaystyle g^{-1}$ are open. and open sets generate the $\displaystyle \sigma$-algebra you are working with).