Denote $f_s(x):=f(sx)$ as the dilation of a function $f$. I want to know whether the following statement is true:

Suppose $f$ and $g$ are measurable functions on $\mathbb{R}$, and $f$ is not almost everywhere equal to $0$. If $gf_s$ is not almost everywhere equal to $0$ for every $s > 0$, then $g$ must be equal to $0$ almost everywhere.

Obviously, this statement must be true if $f$ is continuous. I don't know whether it still holds in the general measurable case (even if $g$ is continuous).

This might seem to be a meaningless question, but it has background. In fact, if $f \in L^2$ and $g$ is a Schwartz function, then from the Plancherel Theorem of Fourier Transform, there is an equivalent statement:

Suppose $f \in L^2(\mathbb{R})$ and $g$ is a Schwartz function, and $f \neq 0$ in $L^2$. If $g*f_s \neq 0$ in $L^2$ for every $s >0$, then $g = 0$.

I believe this statement concerning convolution with dilations is equivalent to the injectivity of the Continuous Wavelet Transform:

Suppose $f \in L^2(\mathbb{R})$ and $g$ is a Schwartz function, and $f \neq 0$ in $L^2$. If for every $s > 0$ and $b \in \mathbb{R}$, $$\int g(x)f(s(x-b))\,dx=0.$$ Then $g$ must be $0$.

I think this is a rather interesting and promising question.

For the Continuous Wavelet Transform, there is a Morlet's wavelet reconstruction formula, as shown in this MSE question: Morlet's wavelet reconstruction formula. However, there seems to be an integrability condition (the finiteness of $B_\psi$) for the formula to be well-defined. I wonder whether the Continuous Wavelet Transform is still injective in the more general case.

I am not familiar with the theories of wavelets. Please correct me if there is any mistake.