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Suppose $f$ is a nonnegative integrable function. (Here, the integrals are Lebesgue integrals.) Is there an elementary way to prove that $$ \lim_{n\to\infty}\int_n^\infty f(x) dx = 0 $$ without using the dominated convergence theorem?

Here is how I think you can prove it with the dominated convergence theorem: Since $f(x)\chi_{[n, \infty)}(x) \to 0$ and $\lvert f(x)\chi_{[n, \infty)}(x) \rvert \leq f(x)$, by the dominated convergence theorem, we have that $$ \lim_{n\to\infty}\int f(x) \chi_{[n, \infty)} = \lim_{n\to\infty}\int_n^\infty f(x) dx = 0. $$

Perhaps there is a way to prove it with the monotone convergence theorem?

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    $\begingroup$ Why does it seem so obvious to you that $f(x)\chi_{[n, \infty)}(x) \to 0$ ? $\endgroup$ – DonAntonio Oct 16 '18 at 17:47
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    $\begingroup$ @DonAntonio I thought that for $x$ fixed, $\chi_{[n,\infty)}(x) \to 0$, and so since $f(x)$ is a fixed real number, $f(x)\chi_{[n,\infty)}(x) \to 0$ as well. $\endgroup$ – measuresproblem Oct 16 '18 at 17:52
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    $\begingroup$ Ok, I see now you meant the limit wrt $\;n\;$, of course, and not wrt $\;x\;$ , as I assumed...Thanks. $\endgroup$ – DonAntonio Oct 16 '18 at 18:05
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You can indeed prove this using the monotone convergence theorem. Let $f_n = f\cdot \chi_{[0,n]}$. Then $0\le f_n \le f_{n+1}$ for each $n$, and $f_n \to f$ pointwise. Hence, by the monotone convergence theorem, $$ \int f_n\,dx \to \int f\,dx. $$ Of course $f \ge f_n$, so the above limit really means $\lim_{n\to\infty}\big(\int f\,dx - \int f_n\,dx\big) = 0$. By linearity of the integral, we have $$ \int f\,dx-\int f_n\,dx = \int f\cdot(1-\chi_{[0,n]})\,dx = \int f\cdot \chi_{(n,\infty)}\,dx = \int_n^\infty f\,dx. $$ (Of course, $0\le \int f_n\,dx <\infty$ for each $n$ since $f$ is integrable, so the subtraction above actually makes sense.) The claim follows by letting $n\to\infty$.

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Let $f$ be integrable and non-negative. Then $A \mapsto \int_A f$ is a finite measure, let's call it $\varphi$. Now, a very fundamental lemma about measures asserts that for a decreasing chain of subsets $A_0 \supset A_1 \supset \ldots$ with $\varphi(A_0) < \infty$ we have $\lim_{n \to \infty} \varphi(A_n) = \varphi(\cap_{n \in \mathbb{N}} A_n)$. In your case, the intersection is empty and we obtain the result.

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    $\begingroup$ It's probably good to note that the proof that $\varphi$ is a finite measure relies on something like the monotone convergence theorem. Nevertheless, I like this answer and I upvoted it :-) $\endgroup$ – Alex Ortiz Oct 17 '18 at 20:49
  • $\begingroup$ Actually, I looked it up in the lecture notes when I learned measure theory and we first proved that $\varphi$ is a measure (directly after the definition of the integral) and used it for the monotone convergence theorem. But of course, a lot of those theorems and lemmas can be arranged in a different order. $\endgroup$ – Paul K Oct 17 '18 at 21:00
  • $\begingroup$ That's really nice, actually. The interconnections of these convergence theorems and the relationships between integrals and measures is rich. Thanks for pointing that out. $\endgroup$ – Alex Ortiz Oct 17 '18 at 21:46
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We can go back to the definition of Lebesgue integral and use the fact that for all positive $\varepsilon$, there exists a function $s$ which is a linear combination of indicator function of set of finite measure such that $f-s$ is non-negative and $\int_{\mathbb R}(f-s)(x)\mathrm dx\lt \varepsilon$. Therefore, it suffices to prove the result when $f$ is an indicator function of a set of finite measure. This follows from the fact that if $A$ has a finite measure, then the sequence $\left(A_n\right)_{n\geqslant 1}$ defined by $A_n:=A\cap \left[n,+\infty\right)$ is non-increasing and $A_1$ has a finite measure.

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