Is an $L^p$ function in an annulus $L^p$ restricted to almost all planes? Let $n\geq3$ and consider the annulus-like domain $A=B(0,1)\setminus B(0,r)\subset\mathbb R^n$.
Take any number $p\in[1,\infty]$.
If $f\in L^p(A)$, is it true that $f|_{P\cap A}\in L^p(P\cap A)$ for almost every two dimensional subspace $P$ of $\mathbb R^n$?
Note that all the spaces $P$ go through the origin, so we are not slicing the domain with parallel planes.
If $A$ is replaced with the ball $B(0,1)$, the claim is false in general because the function can be concentrated near the origin.
For example, the function $f(x)=|x|^{-n+1/2}$ is in $L^1$ but it is not $L^1$ restricted to any proper subspace (intersected with the ball).
The Grassmannian of two dimensional subspaces of $\mathbb R^n$ is a compact, smooth manifold and we can equip it with any Riemannian metric; null sets don't depend on the choice of the metric.
Here are some incomplete ideas of mine for proving this, but I feel there should be a more direct proof or at least some theorems that I could use to justify my arguments:


*

*It seems possible that this could be done sphere by sphere.
For a function $f\in L^p(A)$ we have $f|_{S_r}\in L^p(S_r)$, where $S_r=\partial B(0,r)$ is the sphere, for almost every $r$ (a proof can be found in this MSE question).
It should be enough if we can answer the original question with $A$ replaced with the sphere $S_1$: If we have an $L^p$ function on the sphere, is its restriction to almost every great circle in $L^p$?
Let $T_1S_1$ be the unit tangent bundle of $S_1$, and denote the projection by $\pi:T_1S_1\to S_1$.
For $f\in L^p(S_1)$ we have $\pi^*f\in L^p(T_1S_1)$ (the bundle is locally a product so measurability of $f$ implies that of $\pi^*f$).
The space $T_1S_1$ is nicely foliated by great circles with unit speed parametrization, so a Fubini-type theorem should give the desired result.

*For a continuous function $f:\bar A\to\mathbb R$ we have
$$
\int_A |f|^p=\int_G\left(\int_P K|f|^p\right)dP,
$$
where $G$ denotes the Grassmannian of two-planes and $K:A\to(0,\infty)$ is the radial function $K(x)=c|x|^{n-1}$.
The constant $c\in(0,\infty)$ could be calculated explicitly, but it is irrelevant for the argument.
If one can argue that the same identity must hold for $f\in L^p$ as well, the desired conclusion follows.
 A: As felipeh noted, the problem reduces to $p=1$ by replacing $f$ with $g = |f|^p$. (The case $p=\infty$ should be treated separately.) 
Also, the annulus can be replaced by the sphere $S^{n-1}$. Indeed,   define a function $h$ on the unit sphere by $h(\xi) = \int_r^1 g(t\xi)\,dt$. Then $h\in L^1(S^{n-1})$, with a norm comparable to $\|g\|_{L^1(A)}$ because the factor of $\rho^{n-1}$, coming from integration in spherical coordinates, is between $r^{n-1}$ and $1$ and can be safely omitted. Also, the integral of $h$ over a circle is comparable to the integral of $g$ over the corresponding plane, for the same reason. 
With the above (optional) simplifications in mind, one can proceed as follows. Let $G$ be the space of circles (identified with $2$-planes) equipped with normalized (meaning total mass $1$) Grassmannian measure. Define $F:G\times [0,2\pi]\to S^{n-1}$ so that $F(C,\theta)$ is the point on circle $C$ with polar angle $\theta$, as measured from the point of $C$ with the maximal $x_1$ coordinate, initially moving in the direction of the increasing $x_2$ coordinate. These are pretty arbitrary choices to ensure we have a Borel measurable map. It's not defined on some exceptional circles, but those have zero measure in $G$.
Equip the product space $G\times [0,2\pi]$ with the product $\mu$ of normalized measures on $G$ and $[0,2\pi]$. The key point is that the pushforward of this product measure under $F$ is rotation-invariant, by design of $F$. Therefore, this pushforward is the normalized Lebesgue measure $\lambda$ on $S^{n-1}$, as it's the only normalized rotation-invariant measure on the sphere. 
The definition of pushforward measure implies $\int_{S^{n-1}} h\,d\lambda = \int_{G\times [0,2\pi]} h\circ F\,d\mu$.  Therefore, $h\circ F$ is integrable on $G\times [0,2\pi]$, which implies $h$ is integrable on a.e. circle.

Returning to $p=\infty$: in this case, $|f|$ is bounded outside of a null set $N$. Apply the above to $h=\chi_N$.
A: Normal Human already gave a good answer, but I came up with another one that I would like to share.
It is based on the coarea formula and I tried to make it feel natural, with few arbitrary choices.
As has been noted, it suffices to consider $L^1$ functions, so let $f\in L^1$.
(For $f\in L^p$ we have $|f|^p\in L^1$.)
This argument works for any $p<\infty$.
Consider the set $E=\{(x,v)\in A\times S^{n-1};v\cdot x=0\}$ and the projection $\pi:E\to A$ to the first coordinate.
We can think of this as a bundle over $A$, and it has locally a product structure.
There is a natural Riemannian metric on $E$ (inherited from $\mathbb R^n\times S^{n-1}$) and all fibers $E_x=\pi^{-1}(x)$ have the same measure $|S^{n-2}|$.
Due to the local product structure $\pi^*f:E\to\mathbb R$ is measurable and $\int_E\pi^*f=|S^{n-2}|\int_Af$, so $\pi^*f\in L^1(E)$.
Let $G$ be the Grassmannian of two-planes in $\mathbb R^n$.
Define $F:E\to G$ so that $F(x,v)$ is the unique plane containing $x$ and $v$.
The map $F$ is a smooth surjection and its differential is everywhere surjective.
Therefore the coarea formula for manifolds applied to $\pi^*f$ gives us
$$
\int_E\pi^*f
=
\int_{P\in G}\int_{F^{-1}(P)}\pi^*f(z)\frac1{NJ\,F(z)} \,d\sigma_P(z)\,dP,
$$
where $NJ\,F$ is the normal Jacobian of $F$ and $\sigma_P$ is the measure on the level set $F^{-1}(P)\subset E$.
For any plane $P\in G$ we have $\pi(F^{-1}(P))=P\cap A$.
Due to this and rotational symmetry we get
$$
|S^{n-2}|\int_Af
=
\int_E\pi^*f
=
\int_{P\in G}\int_{P\cap A}f(x)\phi(|x|) \,dx\,dP,
$$
where $dx$ stands for the natural measure on the plane $P$ and $\phi$ is some weight function.
The function $\phi$ could be calculated explicitly, but proper scaling requires that $\phi(t)=t^{n-1}$ up to a multiplicative constant.
(By scaling I mean considering functions $f$ that are supported on thin spherical layers. As the radius of this sphere changes, both sides should scale in the same way.)
Therefore there is a constant $c>0$ so that
$$
\int_Af
=
c\int_{P\in G}\int_{P\cap A}f(x)|x|^{n-1} \,dx\,dP.
$$
In particular, $f|_{P\cap A}\in L^p(P\cap A)$ for almost every $P\in G$.
