TITLE: Analogue of Lebesgue differentiation theorem in Orlicz spaces
QUESTION [5 upvotes]: It is well known that $$\lim\limits_{r\rightarrow 0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}=|f(x)|$$ for almost all $x\in\mathbb{R}^{n}$. Here $\chi_{B(x,r)}$ denotes the characteristic function of the open ball $B(x,r)$. I wonder whether there is an analogue of this property in Orlicz spaces, that is, $$\lim\limits_{r\rightarrow0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}=|f(x)| \tag{$\ast$}$$ for almost all $x\in\mathbb{R}^{n}$ ?
Where $\Phi:[0,\infty)\to [0,\infty)$ is an increasing, continuous, convex function with $\Phi(0)=0$ and
$$
\|f\|_{L_{\Phi}(\mathbb{R}^{n})}:=\inf\{\lambda>0:\int_{\mathbb{R}^{n}}\Phi\left(\frac{|f(x)|}{\lambda}\right)dx\leq 1\}.
$$
It is a generalization of the $L_p$ norm. Indeed, if we take $\Phi(t)=t^p,\,1\leq p< \infty$ we get $\|f\|_{L_{\Phi}}=\|f\|_{L_{p}}$.
I think we should define a maximal function such that
$$
M^{\Phi}f(x)=\sup_{r>0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}
$$
and obtain a weak type inequality for this operator. As an application of this we get the desired equation. Unfortunately, I couldn't do any of these operations.

REPLY [4 votes]: I found the answer of my own question. In the paper "Hencl, Stanislav; Kleprlík, Luděk; Composition of q-quasiconformal mappings and functions in Orlicz-Sobolev spaces. Illinois J. Math. 56 (2012), no. 3, 931–955." The authors proved that the equation ($\ast$) holds provided that $\Phi\in\Delta^{\prime}$, i.e. there exist $C>0$ such that 
$$\Phi(x y)\leq C \Phi(x)\Phi(y)\text{ for every }x,y\geq 0.$$
They also add a remark such that:
"Surprisingly we cannot have ($\ast$) for general Young functions, because the term $\|\chi_{B(x,r)}\|$ does not necessarily scale well for small $r$. For $\alpha<0$ we construct a function $f\in L^q\log^{\alpha}L$ such that the limit in ($\ast$) is infinite everywhere. "