If Lebesgue integral of nonnegative measurable function on $[0,x]$ is one-to-one, is $f$ positive almost everywhere?

Question

Let $f:[0,\infty)\to[0,\infty)$ be measurable. Consider the following function: $$F(x)=\int_{[0,x]}f\,d\mu,$$ where $\mu$ is the Lebesgue measure on $\mathbb{R}$. If $f$ is positive almost everywhere on $[0,\infty)$, then $F$ is strictly increasing. This follows from the fact that for every $\varepsilon>0$ and for any fixed $x\in[0,\infty)$, we have: $$\int_{[x,x+\varepsilon]}f\,d\mu>0,$$ since $[x,x+\varepsilon]$ has positive measure and the set of points in $[x,x+\varepsilon]$ on which $f$ vanishes has measure zero. This shows that $F$ is one-to-one if $\mu\big(\{x\in[0,\infty):f(x)=0\}\big)=0$.

Now suppose that $F$ is one-to-one. Does it follow that $f$ vanishes only on a set of measure zero?

Let $\mathcal{O}$ be the set of points in $[0,\infty)$ on which $f$ vanishes. If $\mathcal{O}$ contains any interval $[a,b]$, then we can write: $$\int_{[0,a]}f\,d\mu=\int_{[0,a]}f\,d\mu+\int_{[a,b]}f\,d\mu=\int_{[0,b]}f\,d\mu,$$ so $F(a)=F(b)$ and $F$ is not one-to-one. However, in general $\mathcal{O}$ need not contain any interval - for example if $\mathcal{O}$ is a fat Cantor set.

Thoughts? If $F$ being one-to-one does not imply that $\mu\big(\{x\in[0,\infty):f(x)=0\}\big)=0$, is there any other necessary and sufficient condition for $F$ to be injective?

Answer 1

There is a simple characterization of the measurable functions $f$ for which $F$ is injective. We use Proposition 19.2.6a from Tao's Analysis II that for any measurable set $\Omega$ and any nonnegative measurable $f:[0,\infty]\to[0,\infty]$, we have: $$\int_\Omega f\,d\mu=0\iff\mu\big(x\in\Omega:f(x)=0\big)=\mu(\Omega).$$

Proposition. Let $f:[0,\infty)\to[0,\infty)$ be measurable, and define: $$F(x)\equiv\int_{[0,x]}f\,d\mu,$$ where $\mu$ is the Lebesgue measure on $\mathbb{R}$. Then $F$ is injective if and only if for every interval $[a,b]\subseteq[0,\infty)$, $\mu\big(\operatorname{supp}f\cap[a,b]\big)>0$.

Proof: First suppose that for each interval $[a,b]\subseteq[0,\infty)$ we have $\mu\big(\operatorname{supp}f\cap[a,b]\big)>0$. Fix $x\in[0,\infty)$ and let $\varepsilon>0$. We have: $$\int_{[0,x+\varepsilon]}f\,d\mu=\int_{[0,x]}f\,d\mu+\int_{[x,x+\varepsilon]}f\,d\mu,$$ so to show that $F$ is injective it suffices to show that the rightmost integral is positive. But since $\operatorname{supp}f\cap[x,x+\varepsilon]$ has positive measure by assumption, the rightmost integral must be positive. This shows that $F(x)<F(x+\varepsilon)$ for each $\varepsilon>0$, so $F$ is strictly increasing, hence injective.

Now suppose that there exists an interval $[a,b]\subseteq[0,\infty)$ for which $\mu\big(\operatorname{supp}f\cap[a,b]\big)=0$. This implies that $\mu\big(x\in[a,b]:f(x)=0\big)=\mu\big([a,b]\big)$. Then similarly writing: $$\int_{[0,b]}f\,d\mu=\int_{[0,a]}f\,d\mu+\int_{[a,b]}f\,d\mu,$$ we find that the rightmost integral is zero, and so $F(a)=F(b)$, which shows that $F$ is not injective. $\square$

Answer 2

Let $A$ be a dense open subset of $\mathbb R$ that is not almost all of $\mathbb R$ in the measure-theoretic sense. For example, $A$ could be the complement of a fat Cantor set. Then the characteristic function of $A$ vanishes on a set of positive measure, namely the complement of $A$. Yet its indefinite integral is strictly increasing.

(I'll make this answer community wiki, because it's essentially part of an earlier answer by Kavi Rama Murthy. That answer was deleted, apparently because it contained an additional statement that wasn't correct.)