This is one of the reasons why, instead of saying that the density function of the normal distribution is
$$
\varphi(x) = \frac1 {\sigma\sqrt{2\pi}} \exp\left(-\frac12 \left( \frac{x-\mu}\sigma \right)^2 \right),
$$
I prefer to say that the normal distribution is (**not** that the **density function** of the normal distribution is)
$$
\frac 1{\sqrt{2\pi}} \exp\left( -\frac12 \left( \frac{x-\mu}\sigma \right)^2 \right) \left( \frac{dx} \sigma \right).
$$
(Here there is no stray $\sigma$ that is not accompanied by an $x$.)

A density function is a density function only with respect to a measure, which in this case is $dx,$ i.e. Lebesgue measure on the real line.

A likelihood function is not accompanied by such a measure. You can multiply a likelihood function by a measure and get another measure, and that is just what is done when you multiply the likelihood by the prior. The prior has a $\text{“}dx\text{”}$ in it; the likelihood function does not.

Moreover, densities are not defined pointwise, whereas likelihood functions are. You can change the value of a density function at an isolated point, or at all points in some set of measure $0$ and the density function you get is exactly equivalent. That is why I write (for example)
$$
f(x)\,dx = \begin{cases} \exp(-\lambda x)(\lambda\,dx) & \text{for } x>0, \\ 0 & \text{for } x<0, \end{cases}
$$
with $\text{“}{>}\text{”}$ and $\text{“}{<}\text{”}$ rather than $\text{“}{\ge}\text{”}$ and $\text{“}{\le}\text{”},$ whereas with c.d.f.s and likelihood functions I am careful about which ones are $\text{“}{>}\text{”}$ and which are $\text{“}{\ge}\text{”}.$

And in some cases if you integrate a likelihood function with respect to some obvious measure you get $+\infty,$ so that function cannot be a probability density function. For example, in
$$
f_\theta(x) \,dx = \begin{cases} \dfrac{\theta\,dx}{(\theta x)^2} & \text{if } x>1/\theta, \\[6pt] \,\,\,0 & \text{if } x<1/\theta, \end{cases}
$$
the function $f_\theta(x)$ is a probability density function, and the likelihood function based on a single observation $x>0$ is
$$
L(\theta) = \begin{cases} 1/\theta & \text{if } \theta \ge1/x, \\[4pt] \,\,\, 0 & \text{if } 0<\theta<1/x. \end{cases}
$$
If you integrate that with respect to Lebesgue measure, you get $+\infty.$