# Why can I not use this alternative, simpler way of showing that $\frac{1}{|a|}f(\frac{x-b}{a})$ is Borel measurable

Let $$X$$ be a real random variable and $$f$$ is its density w.r.t. the Lebesgue measure.

As a background, I was asked to show the density of $$Y:=aX+b$$ exists and is $$g(x):=\frac{1}{|a|}f(\frac{x-b}{a})$$. In order to do that, I eventually need to show that $$\frac{1}{|a|}f(\frac{x-b}{a})$$ is measurable. This is where I struggle to find why my alternative would be wrong.

Correct Answer: it is clear since $$\int_{-\infty}^{c}\frac{1}{|a|}f(\frac{x-b}{a})dx$$ exists that $$g(x):=\frac{1}{|a|}f(\frac{x-b}{a})$$ is measurable on $$(\infty, c], \forall c \in \mathbb R$$. And since $$\{ (\infty, c] |c \in \mathbb R\}$$ is a generator of the $$\mathcal{B}(\mathbb R)$$, therefore $$g$$ is borel measurable. (Is this correct reasoning?)

Proposed Alternative Solution: I would think that an alternative way of showing that $$g$$ is Borel-Measurable is simply stating that $$g$$ is the product of borel-measurable functions $$\frac{1}{|a|}$$ (as a constant function is continuous and therefore Borel-measurable) as well as $$f(\frac{x-b}{a})$$, and therefore the product is Borel-measurable. I have a feeling that this may however not be so simple because $$f(x)$$ being measurable does not imply that $$f(\frac{x-b}{a})$$ is indeed Borel-measurable.

Any ideas, as to why the alternative does not hold?

Your alternative does work. Let $$h(x)=(x-b)/a$$. Since $$f:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal B(\mathbb R))$$ is measurable, and $$h:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal L(\mathbb R))$$ is measurable (since it is Borel measurable, and inverse images of Borel null sets are null), it follows their composition
$$f\circ h:(\mathbb R,\mathcal L(\mathbb R))\to (\mathbb R,\mathcal B(\mathbb R))$$ is measurable.

• Do we indeed know $f$ is Borel? Or is it just Lebesgue measurable? – Robert Israel Dec 12 '18 at 2:16
• @RobertIsrael See edit. Whatever level of measurability you have for $f$, the same will be true of $g$. – Mike Earnest Dec 12 '18 at 4:37
• @RobertIsrael Oh sorry, I mistook you for OP! But to your question, I am not sure and might have misspoke before. – Mike Earnest Dec 12 '18 at 4:38
• Careful: the composition of two Lebesgue measurable functions is not necessarily Lebesgue measurable. See e.g. here – Robert Israel Dec 12 '18 at 12:42
• @RobertIsrael You are right, it is now fixed. – Mike Earnest Dec 12 '18 at 14:43

The Radon-Nikodym theorem tells you that if $$X$$ is a random variable whose distribution is absolutely continuous wrt Lebesgue measure (i.e. $$\mathbb P(X \in A) = 0$$ for all sets $$A$$ of Lebesgue measure $$0$$), then this distribution has a density $$f(x)$$ which is a Lebesgue measurable function, i.e. $$\mathbb P(X \in E) = \int_E f(x)\; dx$$ for all Lebesgue measurable sets $$E$$.

It is clear that if $$a \ne 0$$, the distribution of $$Y = aX+b$$ is also absolutely continuous and its density is what you expect it to be: the change-of-variables formula works for Lebesgue integration.

I don't know why you have to drag Borel into this.

• Thank your for the insight, but we "dragged" Borel into this in our lecture, as we have not used Radon-Nikodym as of yet. And I am specifically looking for an answer on whether my alternative solution stands. And if not, why not. – SABOY Dec 11 '18 at 22:41