Proving a limit using the dominated converge theorem.

Question

Let $\Omega \subset \mathbb{R}^n$ be an open set and consider the usual Lebesgue space $L^p(\Omega)$. Adicionally, consider the space

$$ U(\Omega) = \left\{ f \in L^p(\Omega) \, \colon \, \lim_{r \to 0} \, \sup_{x \in \mathbb{R}^n} \, \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy = 0 \right\} .$$

My goal is to show that $L^p(\Omega) = U(\Omega)$, using the Dominated Convergence Theorem.

My attempt. The inclusion $U(\Omega) \subset L^p(\Omega)$ is direct just by definition of $U(\Omega)$. On the other hand, let $f \in L^p(\Omega)$ be arbitrary. In order to show that $f \in U(\Omega)$, it suffices to guarantee that

$$ \lim_{r \to 0} \, \sup_{x \in \mathbb{R}^n} \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy = 0. $$

For every $x \in \mathbb{R}^n, r > 0$ we have that

$$ \int_{B(x,r) \cap \Omega} |f(y)|^p \, dy = \int_{\mathbb R^n}|f(y)|^p \chi_{B(x,r) \cap \Omega}(y) \, dy. $$

Now, there are two questions that arise:

1. The first question is how to deal with the supremum. I believe that before applying the DCT I should find a way to get rid of the supremum but I've been unable to find one.

2. The second question is related to applying the DCT with limits that tend to $0$.There are some posts on MSE about this (e.g. [$1$], [$2$] and [$3$]). The main reasoning presented in such posts is to use the sequential formulation of the limit to retrieve conclusions. I believe that I understand this reasoning but I have a more informal question: in pratice (say in articles or books, for example), what is the most common way of dealing with the DCT when limits tend to zero? Do the authors explicitly describe the use of the sequential formulation of a limit or do they just present a dominating function and move the limit (tending to $0$) inside of the integral ?

Thanks for any help in advance.

Answer 1

The following is an argument using DCT only:

Fix a sequence $r_n \to 0$. We need to prove $\lim_{n \to \infty} \, \sup_{x \in \mathbb{R}^n} \int_{B(x,r_n) \cap \Omega} |f(y)|^p \, dy = 0$. To do this, it suffices to show that for any subsequence of $r_n$, there is a further subsequence along which the above quantity converges to $0$. That is, it suffices to prove, for any sequence $r_n \to 0$, there is a subsequence $r_{n_k}$ s.t. $\sup_{x \in \mathbb{R}^n} \int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy \to 0$. Since $r_n \to 0$, we may choose a subsequence $r_{n_k}$ s.t. $(2r_{n_k})^n < 2^{-k}$.

Claim: For each fixed $r_{n_k}$, there is a $x_k \in \mathbb{R}^n$ s.t. $\sup_{x \in \mathbb{R}^n} \int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy = \int_{B(x_k,r_{n_k}) \cap \Omega} |f(y)|^p \, dy$.

Proof of claim: The integral $\int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy = \int_{\mathbb{R}^n} |f(y)|^p \chi_{B(x,r_{n_k}) \cap \Omega}(y) \, dy$ is continuous in $x$ by DCT (since, if $x_n \to x$, then $|f(y)|^p \chi_{B(x_n,r_{n_k}) \cap \Omega}(y) \to |f(y)|^p \chi_{B(x,r_{n_k}) \cap \Omega}(y)$ pointwise a.e. and they are all dominated by $|f(y)|^p$, which is integrable). Furthermore, $\int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy = \int_{\mathbb{R}^n} |f(y)|^p \chi_{B(x,r_{n_k}) \cap \Omega}(y) \, dy \to 0$ as $x \to \infty$ by DCT as well (again, if $x_n \to \infty$, then $|f(y)|^p \chi_{B(x_n,r_{n_k}) \cap \Omega}(y) \to 0$ pointwise). Hence, $x \mapsto \int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy$ is a continuous function vanishing at $\infty$, so it must achieve its supremum at some $x_k \in \mathbb{R}^n$.

Now, given the claim, we need to prove $\sup_{x \in \mathbb{R}^n} \int_{B(x,r_{n_k}) \cap \Omega} |f(y)|^p \, dy = \int_{B(x_k,r_{n_k}) \cap \Omega} |f(y)|^p \, dy = \int_{\mathbb{R}^n} |f(y)|^p \chi_{B(x_k,r_{n_k}) \cap \Omega}(y) \, dy \to 0$. Again by DCT, it suffices to show $|f(y)|^p \chi_{B(x_k,r_{n_k}) \cap \Omega}(y) \to 0$ pointwise a.e. We observe that $|f(y)|^p \chi_{B(x_k,r_{n_k}) \cap \Omega}(y) \to 0$ when $y \in \cup_{l \geq 1} \cap_{k \geq l} B(x_k,r_{n_k})^c$ (where the superscript $c$ indicates complement), so it suffices to show $[\cup_{l \geq 1} \cap_{k \geq l} B(x_k,r_{n_k})^c]^c = \cap_{l \geq 1} \cup_{k \geq l} B(x_k,r_{n_k})$ is a null set. Note that $\lambda(B(x_k,r_{n_k})) \leq (2r_{n_k})^n < 2^{-k}$, so $\lambda(\cup_{k \geq l} B(x_k,r_{n_k})) < \sum_{k \geq l} 2^{-k} = 2^{-l+1}$. Thus, $\lambda(\cap_{l \geq 1} \cup_{k \geq l} B(x_k,r_{n_k})) = \lim_{l \to \infty} \lambda(\cup_{k \geq l} B(x_k,r_{n_k})) \leq \lim_{l \to \infty} 2^{-l+1} = 0$. This concludes the proof.

Answer 2

This proof doesn't use the Dominated Convergence Theorem, but it uses the fact that a Lebesgue integral over a set of measure zero is $0$: see the first bullet point of Lebesgue integration: Basic theorems of the Lebesgue integral.

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Set $g(x)=|f(x)|^p[x\in\Omega]\in L^1$, where $[\dots]$ are Iverson Brackets.

Preliminaries

Since $$ \lim_{r\to\infty}\int_{|y|\ge R}g(y)\,\mathrm{d}y=0\tag1 $$ there is an $R_0$ so that $$ \int_{|y|\ge R_0}g(y)\,\mathrm{d}y\lt\epsilon/2\tag2 $$ Define $$ h_k(x)=\int_{B(x,1/k)}g(y)\,\mathrm{d}y\tag3 $$ For $|x|\gt R_0+1$, $B(x,1/k)\subset \{y:|y|\ge R_0\}$. Inequality $(2)$ then implies that $$ h_k(x)\lt\epsilon/2\tag4 $$

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Contradiction

Suppose that the hypothesis is false; that is $$ \lim_{k\to\infty}\sup_{x\in\mathbb{R}^n}h_k(x)\ge\epsilon\gt0\tag5 $$ The limit in $(5)$ says that there is a $K$ so that for all $k\ge K$, there exists an $x_k\in\Omega$ so that $$ h_k(x_k)\ge\epsilon/2\tag6 $$ Inequality $(4)$ then implies that $|x_k|\le R_0+1$. That is, $\{x_k\}$ is an infinite sequence in a bounded set. Thus, there is convergent a subsequence $x_{k_j}$ so that $$ \lim\limits_{j\to\infty}x_{k_j}=x_\infty\in\overline{\Omega}\tag7 $$ For any $r\gt0$, there is a $J$ so that for $j\ge J$, $|x_{k_j}-x_\infty|\le r/2$ and $\frac1{k_j}\le r/2$.

This means that $B(x_{k_j},1/k_j)\subset B(x_\infty,r)$, and therefore, $$ \int_{B(x_\infty,r)}g(y)\,\mathrm{d}y\ge\epsilon/2\tag8 $$ Since $(8)$ is true for all $r\gt0$, $$ \int_{\{x_\infty\}}g(y)\,\mathrm{d}y\ge\epsilon/2\tag9 $$ which is impossible since the Lebesgue measure of $\{x_\infty\}$ is $0$. Thus, the assumption made in $(5)$ is false, and therefore, $$ \lim_{k\to\infty}\sup_{x\in\mathbb{R}^n}h_k(x)=0\tag{10} $$ which is equivalent to $$ \lim_{r\to0}\sup_{x\in\mathbb{R}^n}\int_{B(x,r)\cap\Omega}|f(y)|^p\,\mathrm{d}y=0\tag{11} $$