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Let $f$ be a nonegative measurable function on $\mathbb{R}$. Show that

\begin{equation} \boxed{\lim_{n \to \infty} \int_{-n}^{n} f=\int_{\mathbb{R}} f} \end{equation}

This is from Royden 4th, chapter 4 #34.

I know we can easily do this with the monotone convergence theorem, but what about the Dominated convergence theorem? The statement of the dominated convergence theorem states that the dominating function is integrable, and the obvious candidate for that function would just be $f$ since it clearly dominates $f\chi_{[-n,n]}$. But we do not know that $\int_{\mathbb{R}}f$ has finite integral, so how can we use the theorem?

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    $\begingroup$ If the integral of $f$ diverges over $\Bbb R$, then you can pick an $n_0$ such that $\int_{-n}^nf$ diverges for any $n\gt n_0$ and the equality holds trivially as an equality of extended reals. You can then take $f$ as integrable w.l.o.g after this, and apply DCT. As $f$ is non-negative, its integral cannot diverge in the sense of oscillation, such as the series $1-1+1-1\pm\cdots$, so the integral is either $+\infty$ or finite convergent $\endgroup$
    – FShrike
    Dec 19, 2021 at 10:46

1 Answer 1

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If you are required to avoid Monotone Convergence Theorem you can do the following: We may suppose $\int f=\infty$ since the other case is trivial.Let $T$ be any positive number. There exists a simple function $s$ such that $0 \leq s \leq f $ and $\int s >T$. By elementary properties of a measure you can see that $\int_{-n}^{n} f \geq \int_{-n}^{n} s \to \int_{-\infty}^{\infty} s>T$. Hence, $\int_{-n}^{n} f >N$ for $n$ sufficiently large.

We now have a proof that uses only basic properties of a measure (especially continuity from below of a measure) and DCT.

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  • $\begingroup$ Please endeavor not to answer duplicate questions. Questions which cite a particular text are particularly easy to find using Google. $\endgroup$
    – Xander Henderson
    Dec 19, 2021 at 20:04

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