if $f(x)

Question

Let $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$, and $a\in\mathbb{R}$ such that $\forall x\in[a,\infty]\ f(x)<g(x)<h(x)$. Given that $\exists \int\limits_a^\infty f(x)dx,\ \int\limits_a^\infty h(x)dx$, can I claim that $\exists \int\limits_a^\infty g(x)dx$?

I believe I do, but I'd like to assure I'm not missing something.

$\int\limits_a^\infty f(x)dx = L \Longrightarrow \forall \epsilon > 0\ \exists M \in \mathbb{R}:\ \forall m > M\ \left\vert\int\limits_a^m f(x)dx - L\right\vert < \frac{\epsilon}{2} \Longrightarrow \forall m_1, m_2 > M \left\vert\int\limits_{m_1}^{m_2} f(x)dx\right\vert < \epsilon$

Thus $\lim\limits_{k\rightarrow\infty} \int\limits_k^\infty f(x)dx = 0$. Same applies for $h(x)$. We get $\forall \epsilon > 0 \exists M > a:\ \forall M<m_1<m_2,\ -\epsilon < \int\limits_{m_1}^{m_2} f(x)dx < \int\limits_{m_1}^{m_2} g(x)dx < \int\limits_{m_1}^{m_2} h(x)dx < \epsilon $. This holds for every $m_2 > m_1$, hence $\forall \epsilon > 0 \exists M>a:\ \forall m_1,m_2 >M\ \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert< \epsilon$

If $\int\limits_{a}^{\infty}g(x)dx \leq \epsilon$ diverges, then $\exists \epsilon>0:\ \forall M>a \exists m_1,m_2>M:\ \left\vert\int\limits_{a}^{m_1}g(x)dx - \int\limits_{a}^{m_2}g(x)dx\right\vert > \epsilon \Rightarrow \left\vert\int\limits_{m_1}^{m_2}g(x)dx\right\vert > \epsilon$. But we already showed that there exists such an $M$ such that this isn't true for. Hence, the integral converges.

I considered this after studying about the comparison test for positive functions, and thought one could argue a a more general one. Thanks!

Answer 1

I assume you are using Riemann integration.

The answer is no. Consider the following counter-example.

Denote by $\alpha$ the function $\alpha:\mathbb{R}\rightarrow\mathbb{R}$ defined for every $x \in \mathbb{R}$ as follows. $$ \alpha(x) := \frac{1}{x^2}\mathbb{1}_{\mathbb{R}\setminus[-1,1]}. $$

Now consider the following functions. $$ \begin{align} f &:= -\mathbb{1}_{[-1,1]} + -\alpha, \\ g &:= \mathbb{1}_{[-1,1]\cap\mathbb{Q}}, \\ h &:= 2\mathbb{1}_{[-1,1]} + \alpha. \end{align} $$

Then

• $f < g < h$.

• $f$ and $h$ are Riemann integrable on all of $\mathbb{R}$.

• $g$ is not Riemann integrable on any interval that contains $[0,1]$, as $g$ coincides there with the Dirichlet function.

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Addendum

If $g$ is Riemann integrable on $[a,m]$ for every $m \in [a,\infty)$, then the improper integral $\int_a^\infty g(x)\ dx$ does converge. This can be derived from the Cauchy criterion for convergence of an improper integral in a manner similar to the argument you detailed in your post.