Check if $\sup\left\{\int_J|f(x)|\chi_B(x)\, dx:\lambda(B)\le1/k\right\}\to 0$

Question

Let $J\subset\Bbb R$ be a perfect interval and let $f\in\mathcal L_1(J,E)$. Set

$$R_k:=\sup\left\{\int_J|f(x)|\chi_B(x)\, dx:\lambda(B)\le1/k, B\subset J\right\}\tag1$$

I want to see if it is true that $(R_k)\to 0$.

What I tried: let $(A_j)_j$ be any sequence of sets such that $\lim_{j\to\infty}\lambda(A_j)=0$, then $$ \lim_{k\to\infty}\int_J\chi_{A_k}(x)\, dx=\lim_{k\to\infty}\lambda(A_k)=0\tag2 $$ which implies, by a previous result, that there is a subsequence of $(\chi_{A_k})_k$ such that $(\chi_{A_{k_j}})_j\to 0$ point-wise a.e.

Then for any sequence $(A_j)_j$ such that $\lambda(A_j)\le 1/j$ and $\int_{A_j}|f(x)|dx\ge\int_{A_{j+1}}|f(x)|\, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that

$$\lim_{j\to\infty}\int_{A_{k_j}}|f(x)|=0\tag3,$$

and because the sequence $\Big(\int_{A_k}|f(x)|\, dx\Big)_k$ is monotone then we also find that $\lim_k\int_{A_k}|f(x)|\, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)\to 0$ also.

Two questions here:

1. It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)\to 0$ also", that is, if for each $x\in A$ the statements above holds then it also holds for $\sup A$. This, intuitively, seems not right.

2. If the statement of the title is true, there is a simpler way to show the same result?

Answer 1

Firt of all, you cannot expect that $A_{j+1} \subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.

The statement is equivalent to the absolute continuity of the finite measure $$\mu(A) := \int_A |f(x)| \, dx \quad \quad (A \in \mathcal{B}(I))$$ with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with $$\int_{\{|f|>K\}} |f(x)| \, dx < \varepsilon/2.$$ For any measurable set $A \subset I$ with $\lambda(A) < \epsilon/(2K)$ we find that \begin{align} \int_A |f(x)| \, dx &\le \int_{\{|f|>K\}} |f(x)| \, dx + \int_{A \cap \{|f| \le K\}} |f(x)| \, dx \\ & \le \varepsilon/2 + K \lambda(A) < \varepsilon. \end{align} In particular, we have for all $k \in \mathbb{N}$ with $1/k < \varepsilon/(2K)$ that $R_k \le \varepsilon$. This proves $\lim_{k \rightarrow \infty} R_k =0$.