# the composition of a measurable function with a continuous function is measurable

• Sep 30th 2009, 03:01 PM
Dark Sun
the 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 $(-\infty ,\infty )$, then $g\circ f$ is measurable.

• Sep 30th 2009, 06:18 PM
Dark 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, $g$ is measurable.

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

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

Therefore, since $g$ is measurable and $[(-\infty ,\infty )\sim f(x)]\cup f(x)=(-\infty ,\infty )$, the domain of $g$, and both $f(x)$ and $(-\infty ,\infty )\sim f(x)$ are measurable, then by Problem 21.a., the function $g$ restricted to the co-domain of $f$ must be measurable. Then $g(f(x))=g\circ f(x)$ is measurable since for all $x$, an element in the co-domain of $f$ is defined, and $g$ restricted to these elements is measurable.
• Oct 1st 2009, 07:07 PM
putnam120
Look at the inverse of $g\circ f$. So $f^{-1}\circ g^{-1}(U)$ is measurable (U measurable) since, g is continuous and thus measurable. (pull backs of open sets by $g^{-1}$ are open. and open sets generate the $\sigma$-algebra you are working with).