The expression $f(x)\,dx$ is in some contexts taken to refer to the measure $$A \mapsto \int_A f(x)\,dx \quad \text{for all measurable sets } A.$$
Accordingly, if $f$ is a probability density function then $f(x)\,dx$ is a probability measure, and rather than referring to the probability distribution whose density with respect to Lebesgue measure is $f(x),$ as is often done, one can refer to the probability distribution $f(x)\,dx.$ One advantage, besides terseness, is that instead of talking about the density $$ \frac 1 {\Gamma(\alpha)} \lambda^\alpha x^{\alpha-1} e^{-\lambda x} \qquad \text{for } x\ge0, $$ one can talk about the distribution $$ \frac 1 {\Gamma(\alpha)} (\lambda x)^{\alpha-1} e^{-\lambda x} (\lambda\, dx) $$ and that puts that last $\lambda$ where it properly belongs, with $dx,$ and makes clear what substitution should be used in manipulating integrals. Likewise, the normal distribution is $$ \varphi\left( \frac{x-\mu} \sigma \right) \frac{dx} \sigma \quad \text{where } \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}, $$ putting the last $\sigma$ in its proper place.
So now this question has reminded me of something I find irritating about convolutions. One has \begin{align} f(x) \, dx & = e^{-\mu x} (\mu\,dx) \quad\text{for } x\ge0, \\ g(x) \, dx & = e^{-\nu x} (\nu\,dx) \quad\text{for } x\ge0, \end{align} and one is to convolve the two densities to get the density of the sum of two independent random variables. $$ \int_0^x f(s)g(x-s) \, ds = \mu\nu \int_0^x e^{-\mu s} e^{-\nu(x-s)} \, ds. $$ How are we to locate those two factors $\mu,\nu$ in their proper stations according to the philosophy outlined above?