# Prove for $\epsilon >0$ exists a finite interval $[a,b]$ such that $|\int{f(x)}dx-\int_{a}^b f(x)dx|<\epsilon$

Let $$f:\mathbb{R}\rightarrow\bar{\mathbb{R}}$$ Lebesgue integrable. Prove that for $$\epsilon >0$$ there exists a finite interval $$[a,b]$$ such that

$$\left|\int{f(x)}dx-\int_{a}^b f(x)dx\right|<\epsilon.$$

My attempt: If $$f$$ is integrable on $$[a,b]$$, then for any $$\epsilon > 0$$ there exists $$\delta > 0$$ such that for any measurable set $$D \subset [a,b]$$ with measure $$\mu(D) < \delta$$ we have

$$\left|\int_{a}^b f(x)dx\right|<\epsilon/2.$$

This is where I'm stuck. Can someone help me?

• Apply Lebesgue Dominated Convergence Theorem to $\{\chi_{[-n,n]} f\}_{n \in \mathbb{N}}$. – Sean Haight Jul 18 at 22:56
• I don’t understand.... The first integral is indefinite, so the difference will be a constant of integration plus some other value – gen-z ready to perish Jul 19 at 0:23

You can easily show via dominated convergence that $$\lim_{n\to\infty} \int_{-n}^n f(x)dx=\int f(x)dx.$$ By definition, for a fixed $$\epsilon>0$$, this means that for all $$N$$ large enough, $$\bigg\vert \int f(x)dx-\int_{-N}^N f(x)dx\bigg\vert<\epsilon.$$
If $$A=\bigcup_{n=1}^{\infty}A_n$$ and $$A_1\subset A_2\subset\dots,$$ then $$\int_A f\,\text{d}\mu=\lim\limits_{n\to\infty}\int_{A_n} f\,\text{d}\mu.$$ This is the fundamental property of the Lebesgue integral.