# How to prove that $\lim_{n\to\infty}\int_n^\infty f(x) dx = 0$ when $f$ is integrable.

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?

• Why does it seem so obvious to you that $f(x)\chi_{[n, \infty)}(x) \to 0$ ? Oct 16 '18 at 17:47
• @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. Oct 16 '18 at 17:52
• Ok, I see now you meant the limit wrt $\;n\;$, of course, and not wrt $\;x\;$ , as I assumed...Thanks. Oct 16 '18 at 18:05

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.
• 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 :-) Oct 17 '18 at 20:49
• 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.
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 $$n\to\infty$$.
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.