A guess related to Lebesgue differentiation theorem When I read Lebesgue differentiation theorem, I suddenly have the following conjecture, which I can't prove or find a counterexample. 

Let $f\in L_{\mathrm{loc}}^1(\mathbb{R}^n)$. If 
  $$
\int_{B_r(x)} f(y)dy=0
$$
  holds for any $r\geq 1$ and $x\in \mathbb{R}^n$, then can we say that $f(x)=0$ a.e. ? Please be careful that $r$ is larger than 1, which prevents us from taking advantage of  Lebesgue differentiation theorem. When $n=1$, this seems to be true. 

 A: Fix $n$ and let $\mathcal Z\subseteq L_{\mathrm{loc}}^1(\mathbb{R}^n)$ denote the class of solutions $f$ to your equation
$$
\int_{B_r(x)} f(y)dy=0\text{ for all $r\geq 1$ and $x\in \mathbb{R}^n$.}
$$
$\mathcal Z$ has several nice symmetries:


*

*Isometries. If $f\in\mathcal Z$ and $g$ is an isometry of $\mathbb R^n$, then $g\star f\in \mathcal Z$ where $(g\star f)(x)=f(g^{-1}x)$. This is because the set of spheres of radius $r\geq 1$ is preserved by isometries.

*Partial integrals. If $T$ is a manifold equipped with a measure, and $F:T\times\mathbb R^n\to\mathbb R$ is a measurable function such that $F$ is absolutely integrable on $T\times C$ for compact sets $C\subseteq \mathbb R^n$ and such that $F(t,\cdot)\in \mathcal Z$ for almost every $t\in T$, then by Fubini's theorem $f\in\mathcal Z$ where $f$ is the partial integral defined by $f(x)=\int_{t\in T} F(t,x)dt$.

*Convolution by bounded compactly supported functions $\psi:\mathbb R^n\to \mathbb R$. This is a type of partial integral $F(t,x)=f(x-t)\psi(t)$.

*Averaging over the orthogonal group $O(n)$. This is a partial integral $\bar{f}(x)=\int_{g\in O(n)}(g\star f)(x)$. Note this preserves smoothness - the derivatives are just a similar integral over $O(n)$ but with the direction of the derivative varying with $g$.


Therefore, given a function $f\in\mathcal Z$ that is not almost everywhere zero:


*

*We can assume $f$ is smooth by convolving with a compactly supported mollifier. (The usual Lebesgue differentiation theorem can be used to show this doesn't result in the zero function.)

*We can then assume $f(0)\neq 0$ by translating.

*We can then assume that $f$ is radially symmetric by averaging over the orthogonal group, which replaces $f(x)$ by the average of $f$ over the sphere of radius $|x|$.


So $f$ is a smooth radially symmetric function. For $r>1$, the function $I(r)=\int_{B_r(0)}f(y)dy$ is zero, so its derivative $I'(r)=\int_{S_r(0)}f(y)dy$ is zero, where $S_r(0)$ denotes the sphere of radius $r$ around $0$. But $f(y)$ is constant in $S_r(0)$, so
$$f(x)=0\text{ for all $|x|>1$.}$$
Then:


*

*If we can show that the Radon transform of $f$ is zero, then $f$ is zero. So pick a hyperplane $H=\{x\mid (x\cdot n)=c\}$. Consider the sphere $S_R(n(c+R))$ tangent to $H$ at $nc$, with $R$ large - it is geometrically clear that the intersection of $S_R(n(c+R))\cap B_1(0)$ is approximately the same as $H\cap B_1(0)$. Since $f$ is continuous, $\int_{S_R(n(c+R))} f \to \int_{H} f$ as $R\to\infty$. We showed before that the integral of $f$ on a sphere of radius at least $1$ is zero, so $\int_{H} f$ is zero as required.

*Alternatively, use the fact that $f$ is in $L_1$ so its Fourier transform is a function $\hat f\in L_\infty$, not almost everywhere zero. The Fourier transform $\widehat{\chi_{B_1(0)}}$ of the indicator function of the unit ball is almost everywhere non-zero; see this explicit description or use Schwarz's Paley-Wiener theorem. But if $f\in\mathcal Z$, then the convolution of $f$ by $\chi_{B_1(0)}$ is everywhere zero, so by the $L_1$ convolution theorem the product $\hat f\widehat{\chi_{B_1(0)}}$ is everywhere zero, which implies $\hat f$ is almost everywhere zero, a contradiction.
A: For $n=1$ it is true.

Claim: Suppose $f\in L^1_{loc}(\mathbb{R})$ satisfies $$\int_{x-r}^{x+r}f(y)dy=0$$ for all $x\in\mathbb{R}$ and all $r\ge1$. Then
  $$\int_{x-r}^{x+r}f(y)dy=0$$ for all $x\in\mathbb{R}$ and all $r>0$.

Proof: Let $x\in\mathbb{R}$ and $r>0$. Then
$$0=\int_{x+r-2R}^{x+r}f(y)dy - \int_{x-r-2R}^{x-r}f(y)dy = \int_{x-r}^{x+r}f(y)dy - \int_{x-r-2R}^{x+r-2R}f(y)dy$$
for all $R\ge1$. By taking different choices of $R$, it follows immediately that
$$\int_{x'-r}^{x'+r}f(y)dy$$
is independent of $x'\in\mathbb{R}$, and therefore it must be $0$ by the assumption. The statement follows.

I don't know if a similar proof can work for $n\ge2$. If we were able to work with squares instead of balls, then it should, but with balls we get "moons" which are difficult to compare.
A: Suppose $n=1.$ For $x>0,$ define$F(x)=\int_{-2}^x f(t)\,dt.$ By hypothesis, $F(x) = 0$ for $x>0.$ Therefore $F'(x) = 0$ on $(0,\infty).$ But by the LDT, $F'(x) = f(x)$ for a.e. $x\in \mathbb R.$ Thus $f(x)=0$ for a.e. $x>0.$ Apply this to $f(-x)$ to then see $f(x)=0$ for a.e. $x\in \mathbb R.$  
A: I am not sure if this is correct, but maybe the idea could work. Fix $x,y\in\mathbb{R}^n$ with $d=|x-y|\geq2$. Let $z=(x+y)/2$ be the midpoint. Let $z_\delta^+=x+(1+\delta)(y-x)/2$ and $z_\delta^-=x+(1-\delta)(y-x)/2$, for $\delta>0$ small. By hypothesis, $$0=\int_{B(z_\delta^+,d/2)} f-\int_{B(z_\delta^-,d/2)} f=\int_{A_\delta^+} f-\int_{A_\delta^-} f,$$ where $$A_\delta^+=B(z_\delta^+,d/2)\backslash (B(z_\delta^+,d/2)\cap B(z_\delta^-,d/2))$$ and $$A_\delta^-=B(z_\delta^-,d/2)\backslash (B(z_\delta^+,d/2)\cap B(z_\delta^-,d/2)).$$ Let $A_\delta=|A_\delta^+|=|A_\delta^-|$. We have $$\frac{1}{A_\delta}\int_{A_\delta^+} f=\frac{1}{A_\delta}\int_{A_\delta^-} f.$$ Letting $\delta$ go to $0$, by a version of the Lebesgue differentiation theorem, we obtain $f(x)=f(y)$. Then $f$ is constant, so $f=0$.
