Product of a $L^{p}$ function and measure

Question

I read somewhere that the product of an $L^{p}$ function for $p \in [1, \infty]$ and a measure is not well-defined. But why?

If we have a function $f$ and measure $\mu$ both defined on $\mathbb{R}$ then usually we can define the product $f \mu$ as a measure $$ (f \mu)(A) := \int_{A}f ~d\mu$$ for any $A \subset \mathbb{R}$.

I don't immediately see the problem with $f$ being $L^{p}$. Is it because $f$ is only defined a.e. with respect to the Lebesgue measure? Then maybe the zero measure sets for $\mu$ and the Lebesgue measure do not necessarily coincide? i.e. a zero Lebesgue measure set where $f$ is not defined may actually be a positive measure set with respect to $\mu$ and therefore the integral over that set wouldn't make sense.

Answer 1

Your intuition is correct. You cannot multiply a 'function' in $L^p(\nu)$ with a measure $\mu$ unless $\nu$ is absolutely continuous w.r.t. $\mu$. This is independent of summability issues and just a consequence of the fact that $L^p(\nu)$ is a space of classes.

To better understand, denote by $\mathcal L^p(\nu)$ the space of measurable functions and by $L^p(\nu)$ the space of classes of elements in $\mathcal L^p(\nu)$. For the sake of clarity $\mathcal L^\infty$ is the space of bounded (not: $\nu$-essentially bounded) measurable functions.

The functional $\| f \|_p:=\big(\int |f|^p d\nu\big)^{1/p}$ is a semi-norm on $\mathcal L^p(\nu)$. It defines an equivalence relation on $\mathcal L^p(\nu)$ by declaring that two elements $f,g$ are equivalent if $\|f-g\|_p=0$. The space $L^p(\nu)$ is the quotient space of $\mathcal L^p$ by this equivalence relation, and the semi-norm $\|\cdot\|_p$ on $\mathcal L^p(\nu)$ descends to a true norm on $L^p(\nu)$ denoted by the same symbol.

Now, let us assume that $\nu$ is a probability measure. In this way $L^\infty(\nu)\subset L^p(\nu)$ for every $p\in [1,\infty]$, and providing a counterexample for $p=\infty$ provides one as well for every $p\in [1,\infty)$.

In order for a multiplication $L^p(\nu)\times \mathcal M\to \mathcal M$ to be well-defined, it must be independent of the representative that you choose for a class in $L^p(\nu)$. This is certainly not the case if $\mu\not\ll\nu$.

For example, let $\mu=\delta_0$ and $\nu$ be the Lebesgue measure on $[0,1]$. For any class $[f]\in L^\infty(\nu)$ (hence in $L^p(\nu)$ for every $p$), both $f$ and $f+\chi_{\{0\}}$ are representatives of $[f]$, where $\chi_{\{0\}}$ is the characteristic function of the singleton $\{0\}$. Thus, $[f] \cdot \mu$ is well-defined only if $$f\cdot \mu= (f+\chi_{\{0\}})\cdot \mu.$$ Clearly this is not the case, since $$f(0)=(f\cdot \mu)\{0\}\neq f(0)+1=\big((f+\chi_{\{0\}})\cdot\mu\big)\{0\}.$$

Even if $\mu\ll\nu$, then you still have the summability issues specified in the comments to your question. Thus, also in this case it might still not be possible to multiply $f\in L^p(\nu)$ by $\mu$ and obtain a measure in the standard sense.