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Let $f_{m,n}(x)$ be a sequence (dependent on $m$, $n$) of Lebesgue integrable functions on $\mathbb{R}$.

Suppose that $f_{m,n}(x)\to 0$ as $m,n\to+\infty$, for almost $x\in\mathbb{R}$; in addition, $\left|f_{m,n}(x)\right|\le g(x)$ for all $m,n\in\mathbb{N}$, for all $x\in\mathbb{R}$, where $g\in L^1(\mathbb{R})$.

Can we apply the Lebesgue dominated convergence theorem to conclude that $\int_\mathbb{R} {{f_{m,n}}\left( x \right)dx} \to 0$ as $m,n\to+\infty$?

(Here notice that $a_{m,n}\to 0$ iff for all $\forall\varepsilon>0$, $\exists N\in\mathbb{N}$: $\forall m,n\ge N\Longrightarrow|a_{m,n}|<\varepsilon$).

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Yes. You can note that you can see $\{f_{n, m}\} $ as a single index sequence. Or you can check that the proof of the dominated convergence theorem applies to your situation.

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  • $\begingroup$ I do not know more when you said that you can see $f_{n,m}$ as a single index sequence. Can you explain in more detail for me? $\endgroup$ – Cao Oct 14 '14 at 10:17
  • $\begingroup$ The only reason why the statement of the DCT does not apply directly to your situation is that you have a doubly indexed sequence. But it is still a sequence, so the theorem does apply. $\endgroup$ – Martin Argerami Oct 14 '14 at 10:27
  • $\begingroup$ Yes. Thank you so much. $\endgroup$ – Cao Oct 14 '14 at 10:31
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Suppose that $\int f_{m,n}$ does not converge to $0$. Then there are sequences $m_k$, $n_k$ tending to infinity such that $\int f_{n_k,m_k}$ does not converge to $0$. Now apply Lebesgue to the single index sequence $g_k=f_{n_k,m_k}$ and obtain a contradiction.

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  • $\begingroup$ Yes. I understood your answer. Thank you very much. $\endgroup$ – Cao Oct 14 '14 at 10:21

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