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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 $\theta$.

Intuition: I think the claim is true because the condition is true for every $\theta$ and thus can be thought as a "scanning condition". I.e. we are scanning $g$ over the whole interval $[0,M]$ by moving the $\theta$'s. Moreover, the density is always strictly positive so this makes the "scanning" meaningful.

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!

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    $\begingroup$ A function of bounded variation can fluctuate infinitely many times from above to below $0$, e.g. $f:[0,1]\to\mathbb{R}$, $f(0)=0$ and $f(x)=x^2\sin(1/x)$ for $x>0$. In fact for that same function and fixed $\theta\in(0,1]$ it should not be difficult to find a measure $\mu_\theta$ as you say so that $\int g(z)d\mu_\theta(z)=0$ right? $\endgroup$
    – Saúl RM
    Oct 30, 2022 at 15:33
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    $\begingroup$ I think it could be helpful if you share your proof, even assuming that $g$ has only finitely many fluctuations from above to below zero. $\endgroup$ Oct 30, 2022 at 17:39
  • $\begingroup$ @SaúlRM I agree with you. My fault. When I requested bounded variation I was really trying to avoid cases like the topologist’s sine, but I forgot about the existence of the case that you mention. Indeed the function you propose is of bounded variation but has infinitely many fluctuations. My idea of proof works only if we exclude that case. I will share it later. How is it called the class of functions that cannot have infinitely many fluctuations? $\endgroup$ Oct 30, 2022 at 18:21
  • $\begingroup$ Sketch of proof: suppose $g$ Is not the zero function, is continuous and has finitely many fluctuations above and below zeros. Consider the biggest neighbour of $0 \in B$ such that $g > 0$ (could be also negative, doesn’t change reasoning). If $g$ is zero around 0, then take the neighbourhood starting from the point where $g$ is nonzero. Such a neighbourhood exists as $g$ is continuous and nonzero. Then $B$ will have the form $B=[0, y)$. Then $\int_{0}^{y} g(z) d \mu_y (z) > 0$ which is not possible. Thus $g$ must be zero up to $y$. Repeat this procedure for the finitely many oscillations. $\endgroup$ Oct 30, 2022 at 19:08
  • $\begingroup$ Your sketch of proof is incomprehensible to me overall, as is almost any part of it. $\endgroup$ Oct 30, 2022 at 19:11

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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 $\int_0^\theta g(z) f_{\theta,s,t}(z) \, dz=0$.

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    $\begingroup$ Thank you a lot! $\endgroup$ Oct 30, 2022 at 20:07
  • $\begingroup$ However, I think that if the family of measures is fixed as this one: $f_{\theta} (z) = \frac{\mathbb{1}_{[0,\theta)}(z)}{2\sqrt{\theta^2 - \theta z}}$ then isn't possible to find $g \neq 0$ independent of $\theta$ s.t. $0 = \int_{0}^{\theta} g(z) d \mu_{\theta} (z) = \int_{0}^{\theta} g(z) \underbrace{f_{\theta}(z)}_{> 0} dz \: \: \: \: \: \: \: \forall \: \theta \in [0,M]$. The reason is that one may try to construct a $g$ that fluctuates infinitely many times as the measures that you proposed, $\endgroup$ Oct 30, 2022 at 22:44
  • $\begingroup$ but we should still be able to find $ \theta^{\prime} \: \: : \: \: 0 \neq \int_{0}^{\theta^{\prime}} g(z) f_{\theta^{\prime}}(z) dz$. $\theta^{\prime}$ is contained in the interval where the function $g$ fluctuates infinitely. This is because $g$ must be independent of $\theta$. $\endgroup$ Oct 30, 2022 at 22:45
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    $\begingroup$ Btw even if $g$ is integrable respect to Lebesgue measure, it may not be integrable respect to $\mu_\theta$. If $g$ is bounded that's not a problem though. If you want check that if $\int g\cdot f_\theta dz=0$ for all $\theta$, then $g=0$, it would be enough to check that the subspace of $L^1([0,M])$ generated by the functions $f_\theta$ is dense in $L^1$. I don't know if that is true for your previous example though, to prove things like that the only tool I know is the Stone Weierstrass theorem and it doesn't seem to work here $\endgroup$
    – Saúl RM
    Oct 31, 2022 at 0:54
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    $\begingroup$ Whether the $f_\theta$ from your example generate a dense subspace of $L^1$, or equivalently, if the functions $g_\theta(z)=\frac{1_{(0,\theta)}}{\sqrt{\theta-z}}$ generate a dense subspace of $L^1$, looks like it could be true but I'm not sure how to prove it. Maybe it's worth asking as a separate question $\endgroup$
    – Saúl RM
    Oct 31, 2022 at 1:05

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