Strong convergence in $L^1$, $\mathbb{R}^d$ with $d>1$ A sequence of positive function $f_n(x)\rightarrow f(x)$ pointwise on the unit ball $B:=\{x:|x|\leq 1\}\subset \mathbb{R}^d$ as $n\rightarrow \infty$. Suppose the sets $\Lambda_t^n=\{x:f_n(x)\geq t\}$ satisfy the following two properties. The Lebesgue measure of their boundary is zero $\mu(\partial \Lambda_t^n)=0$ for all $n\in \mathbb{N}$ and $t\in \mathbb{Q}$, and for all $n,m\in \mathbb{N}$ we have $\Lambda_m^n\subset \{x:|x-x^n_m|\leq 1/m\}$ for some $x^n_m\in B_1(0)$. Prove that $f_n\rightarrow f$ strongly in $L^1(B_1)$ if the dimension $d>1$.
I managed to show that $f_n$ and $f$ are measurable, but I have no idea how to proceed. I don't know what is special about $\mathbb{R}$ that makes it different from $\mathbb{R}^d$ with $d>1$. Any help would be appreciated.
 A: By assumption, for fixed $n$, we have $f_n (x) \leq 1 + \sum_{k = 1}^\infty \chi_{\Lambda^n_k(x)}$ for all $x \in B_1(0)$, it follows that $$\int f_n d{\mu} \leq \mu\left( B_1(0)\right)+ \sum_{k = 1}^\infty \mu\left( \Lambda_t^n\right) \leq  \mu\left( B_1(0)\right)+ \sum_{k = 1}^\infty \mu\left(B_{1/k}(x^n_k)\right) \leq  \mu\left( B_1(0)\right)+ \sum_{k = 1}^\infty \left(\frac{2}{k} \right)^d.$$ Therefore, for $d > 1$, we have $f_n \in L^1(B_1)$.
By Fatou's Lemma, we have $$\int_{B_1} f d{\mu}=  \int_{B_1} \liminf_{n \to \infty} f_n d{\mu} \leq \liminf_{n \to \infty} \int_{B_1} f_n d{\mu}\leq   \mu\left( B_1(0)\right)+ \sum_{k = 1}^\infty \left(\frac{2}{k} \right)^d,$$ it follows that $f$ is also in $L^1(B_1)$. Fix any $\epsilon > 0$. Since $B_1$ is a finite measure space, there is $\delta_1 > 0$ such that $\mu(F) < \delta_1$ implies $\int_F f d{\mu} < \epsilon$. Find $N \in \mathbb{N}$ such that $\sum_{k = N}^\infty (2/k)^d < \epsilon/2$. Pick $\delta_2 > 0$ sufficiently small so that $\delta_2 \leq \mu(B(0,1/N))$ and $N \delta_2 < \epsilon/2 $. Then for any $n$ and $\mu(F) < \delta_2$, take $G = \left\lbrace x; f_n(x) > N\right\rbrace$, we have $$\int_F f_n d{\mu} = \int_{F\cap G} f_n d{\mu} + \int_{F \setminus G} f_n d{\mu} \leq \epsilon/2 +  \mu(F)N  $$ \par
Take $\delta = \min(\delta_1, \delta_2)$. By  Egorov Theorem,  there is measurable $A \subset B_1(0)$ such that $\mu \left( B_1(0) \setminus A\right) < \delta$ and $f_n \to f$ uniformly on $A$.  Then we have [ 
\begin{split}
\lim_{n \to \infty}\int_X |f - f_n|d{\mu} &= \lim_{n \to \infty}\int_{A} |f - f_n|d{\mu} +\lim_{n \to \infty} \int_{X \setminus A} |f - f_n| d{\mu}\\ & \leq \lim_{n \to \infty}\int_{A} |f - f_n| d{\mu} + \limsup_{n \to \infty}\int_{X \setminus A} f d{\mu} +\limsup_{n \to \infty}\int_{X \setminus A} f_n d{\mu} \leq 2 \epsilon
\end{split}
] Since $\epsilon$ is arbitrary, this shows that $f_n \to f$ strongly in $L^1(B_1)$ for $d > 1$. For the counterexample when $d = 1$, define $f_n = \sum_{k = 1}^n \chi_{B(0,1/k)}$. Then $f = \sum_{k = 1}^\infty \chi_{B(0,1/k)}$, but it's clear that $f$ is not integrable; hence $f_n$ doesn't converges to $f$ strongly in $L^1(B_1)$. 
