Continuity and measurability

Question

this question concerns continuity and measureability.

Am I right in thinking that if $f>0$ for all x then $\log(f(\lambda x)/f(x))$ is a continuous function for all $x$. Does this then mean that $\lim_{x\rightarrow\infty}\log\left(\frac{f(\lambda x)}{f(x)}\right)=\log\left(\lim_{x\rightarrow\infty}\frac{f(\lambda x)}{f(x))}\right)$? I know that $f$ is measurable if this is any use. Sorry I forgot to say that $\lambda>0$.

Hmmm does the fact that $\frac{f(\lambda x)}{f(x)}\rightarrow 1$ uniformly in $\lambda$ mean that $\lim_{x\rightarrow\infty}\log\left(\frac{f(\lambda x)}{f(x)}\right)=\log\left(\lim_{x\rightarrow\infty}\frac{f(\lambda x)}{f(x))}\right)$?

Thanks for any help!

Answer 1

No, if $\lambda = 2$ and you take $f:\mathbb{R} \to \mathbb{R}$ with $f(x) = 2$ if $x \in \mathbb{R}\setminus\{2\}$ and $f(2) = 1$, then you have a measurable $f>0$.

$\ln\frac{ f(2x)}{f(x)} = \ln \frac{2}{2} = \ln 1 $ if $x \in \mathbb{R}\setminus\{1,2\}$, $\ln\frac{ f(2(1))}{f(1)} = \ln \frac{1}{2} $ and $\ln \frac{f(2(2))}{f(2)} = \ln\frac{2}{1} = \ln 2$. So the composition is not continuous at $1$ and $2$.

Answer 2

If you'd like a really pathological counterexample, consider $f(x)=\chi(x) + 1$ where $\chi$ is the characteristic function of the rationals. Then $f$ is positive, (Lebesgue) measurable, and nowhere continuous.

Your transformation will result in another everywhere discontinous function, so your limits won't exist.

Edit: Assuming $\lambda$ irrational.