Deduce that a function is zero on interval $[0,M]$

Question

I have been thinking about this for the last few days but I was not able to produce a definitive answer.

Take an integrable function $g$ that maps in $\mathbb{R}$ and with domain contained in $[0,M]$ (not necessarily equal to this interval). Consider now a family of probability measures $\mu_{\theta}$ such that $\mu_{\theta} \ll \lambda$ (Lebesgue measure) for every $\theta$. These measures have strictly positive density $f_{\theta}(z) > 0 \: \: \forall \: 0 \leq z < \theta \leq M$. The densities are not defined for $z \geq \theta$. In symbols:

$$ \mathcal{M}_{\lambda} = \bigg\{ \mu_{\theta} \: : \: \theta \in [0,M] \: \: \text{and} \: \: \frac{d\mu_{\theta}}{d\lambda} = f_{\theta} \: : \: f_{\theta}(z) > 0 \: \: \forall \: 0 \leq z < \theta \leq M, \:\forall \: \theta \in [0,M] \bigg\}$$

Question: is it true that:

$$ 0 = \int_{0}^{\theta} g(z) d \mu_{\theta} (z) = \int_{0}^{\theta} g(z) \underbrace{f_{\theta}(z)}_{> 0} dz \: \: \: \: \: \: \: \forall \: \theta \in [0,M] \: \: \: \: \: \: \stackrel{?}{\implies} \: \: \: \: \: \: g(z) = 0 \: \: \: \: \: \: \: \text{a.e. for} \: z \in [0,M]$$

As an example, you can take as $f_{\theta}(z) = \frac{\mathbb{1}_{[0,\theta)}(z)}{2 \sqrt{\theta^2 - \theta z}}$ for fixed $\theta$.

Proposal: I have at the moment a proof of this statement that assumes that $g$ is continuous and have finitely many zeros. Basically I use the finitely many zeros assumption to say that $g$ cannot have infinitely many fluctuations from above to below zero and then I consider each one of these finitely many neighborhoods where $g$ is strictly positive or strictly negative (by continuity) and I show via contradiction that on that neighborhood actually $g$ must be $0$ because otherwise, we can find a $\theta$ in the middle of a such neighborhood so that the integral is nonzero. Therefore I would like to know if the claim holds more generally for not continuous functions. Or for all continuous functions, not necessarily with finitely many zeros.

Any help is extremely appreciated!

Answer 1

As you say, it is not difficult to prove that $g=0$ when there are finitely many fluctuations in sign.

More precisely, suppose we have points $0=a_0,a_1,\dots,a_n=M$ such that $g\geq0$ or $g\leq0$ in each interval $[a_i,a_{i+1}]$. Then, if we have measures $\mu_\theta$ as you say and $\int_0^\theta g(z) \, d \mu_\theta (z)=0$ for all $\theta$, then we can prove by induction on $i$ that $g=0$ almost everywhere in $[a_i,a_{i+1}]$.

More in general, if the function $g$ satisfies that for every $x\in[0,M]$ there is some $\varepsilon>0$ such that $g\geq0$ or $g\leq0$ in $[x,x+\varepsilon]$, we can prove in the same way that if $\int_{0}^{\theta} g(z) d \mu_{\theta} (z)=0$ for all $\theta$, then $g=0$ (consider the maximum $x$ such that $g\neq0$ almost everywhere in $[0,x]$, and if $x<M$ obtain a contradiction).

If $g$ is not as in the previous paragraph (this of course implies $g\neq 0$), then we can find some measures $\mu_\theta$ as in the question so that $\int_0^\theta g(z) \, d \mu_\theta (z)=0$ for all $\theta\in[0,M]$.

To do it, let $k=\inf\{x\in[0,M];\not\exists\varepsilon>0\text{ such that } g\geq0\text{ or }g\leq0\text{ in }[x,x+\varepsilon]\}$. Then we can prove as before that $g=0$ a.e. in $[0,k]$. Moreover, by definition of $k$, for each $n\in\mathbb{N}$ there are sets of positive measure $A_n,B_n$, such that $A_n\subseteq[k,k+\frac{1}{n}],B_n\subseteq[k,k+\frac{1}{n}]$, $g>0$ in $A_n$ and $g<0$ in $B_n$.

Now let's create the measures $\mu_\theta$: for $\theta\leq k$ we can choose any measures we want. For $\theta>k+\frac{1}{n}$, consider for each $s,t>0$ the density functions $f_{\theta,s,t}=s\chi_{A_n}+t\chi_{B_n}+\chi_{[0,\theta]\setminus(A_n\cup B_n)}$.

Then $\int_0^\theta g(z) f_{\theta,s,t}(z) \, dz = \int_{[0,\theta] \setminus(A_n\cup B_n)} g(z) \, dz + s\int_{A_n}g(z) \, dz+t\int_{B_n}g(z) \, dz$. As $\int_{A_n}g(z) \, dz>0$ and $\int_{B_n}g(z) \, dz<0$, you can adjust $s,t$ so that $\int_0^\theta g(z) f_{\theta,s,t}(z) \, dz=0$.