Cauchy criterion for improper integral

Question

I am studying the convergence of improper integrals, with the following definition:

If $f\colon\mathbb{R}\longrightarrow\mathbb{C}$ be a Riemann integrable function on each $[N,M]$ interval, and
\begin{equation} \int_{-\infty}^{\infty}|f(x)|dx=\lim_{\substack{M\longrightarrow+\infty,N\longrightarrow-\infty}}\int_{N}^{M}f(x)dx, \end{equation}

We will say that f is absolutely integrable.

Regarding this definition, I have several doubts.

1. I have found that $\int_{-\infty}^{\infty}f(x)dx$ there exists if, For all $\varepsilon>0$ there exists $M_{0}$, such that

\begin{equation} \left|\int_{-N}^{-M} f(x) \, dx\right| + \left|\int_{N}^{M} f(x) \, dx\right| < \varepsilon \end{equation} for all $M,N\geq M_{0}.$

2. If $\int_{-\infty}^{\infty}|f(x)|dx$ there exists, then $\int_{-\infty}^{\infty}f(x)dx$.

In the proof, it is suggested to show that

\begin{equation} \int_{-N}^{-M} |f(x)|dx \, dx + \int_{N}^{M} |f(x)| \, dx < \varepsilon \end{equation} is Cauchy. However, I have considered taking $a_{n}=\int_{n-M_0}^{n+M_0}f(x)dx$, But I am unsure.

3. $f$ is absolutely integrable, there exists $M>0$ such that $\int_{|x|>M}|f(x)|dx<\varepsilon$. When I can use this definition?, I think it is only for positive functions, however I am not sure

I'm uncertain about how to establish the connection between the improper integral and condition 2.

Answer 1

Under the conditions given at the beginning of the OP, the improper integral exists if $I=\lim_{M,N\rightarrow\infty}\int^M_{-N}f$ exists. When this happens, $\int^\infty_{-\infty}f:=I$.

The Cauchy principle states that $\int^\infty_{-\infty}f$ converges (i.e., exists) iff for any $\varepsilon>0$, there is $a_\varepsilon>0$ such that for any $M>N>a$ and $M'>N'>a$ $$\Big|\int^{-N'}_{M'}f+\int^M_Nf\Big|<\varepsilon$$

Necessity is simple and I leave it to the OP to prove it. Sufficiency is based n the fact that the space $\mathbb{R}$ of real numbers $\mathbb{R}$ with the metric $d(x,y)=|x-y|$ is a complete space, i.e. every Cauchy sequence in $\mathbb{R}$ converges in $\mathbb{R}$. There are many proves of this in Calculus text books. I provide a prove at the end of this posting.

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Connection between statements (1) and (2) in the OP: The hypothesis of condition (2) means, in terms of Cauchy principle, that for any $\varepsilon>0$, there is $a_\varepsilon>0$ such that for all $M>N>a$ and $M'>N'>a$, $$\Big|\int^{-N'}_{-M'}|f|+\int^M_N|f|\Big|=\int^{-N'}_{-M'}|f|+\int^M_N|f|<\varepsilon$$

The triangle inequality along with the fact that $\big|\int_If\big|\leq\int_I|f|$ an any bounded closed interval $I$ (which the assumption of Riemann integrability on any such interval $I$ guarantees) implies that \begin{align} \Big|\int^{-N'}_{-M'}f+\int^M_Nf\Big|&\leq \Big|\int^{-N}_{-M}f\Big|+\Big|\int^M_Nf\Big|\\ &\leq \Big|\int^{-N'}_{-M'}|f|+\int^M_N|f|\Big|\\ &=\int^{-N'}_{-M'}|f|+\int^M_N|f|<\varepsilon \end{align} Cauchy's principle that implies that the convergence of the improper integral $\int^\infty_{-\infty}|f|$ implies the convergence of the integral $\int^\infty_{-\infty}f$.

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On the equivalence of Statment (3) is equivalent to (1) with $|f|$ in place if $f$: First notice that for numbers $m'>n'>a>0$ and $m>n>a$, $[-n',-m']\cup[n,m]\subset (-\infty,-a)\cup(a,\infty)$. As $|f|\geq0$ $$\int^{-m'}_{-n'}|f|+\int^m_n|f|\leq \int^{-a}_{-\infty}|f|+\int^\infty_a|f|=\int_{\{|x|>a\}}|f(x)|\,dx$$ This shows that (3) implies the Cauchy condition for the the improper integral of $\int^\infty_{-\infty}|f|$

Conversely, the Cacuhy property for the improper integral $\int^\infty_{-\infty}|f|$ implies that that for any $\varepsilon>0$, there is $a'>0$ such that if and $M>N>a'$ $$ \int^{-N}_{-M}|f|+\int^M_N|f|<\frac{\varepsilon}{2}$$ Fix $N>a$. Since $|f|$ is positive, the map $I(M):M\mapsto\int^{-N}_{-M}|f|+\int^M_N|f|$ is nondecreasing and bounded below; hence $\lim_{M\rightarrow\infty}I(M)$ exists and $$\lim_{M\rightarrow\infty}I(M)=\int_{\{|x|>N\}}|f|=\lim_{M\rightarrow\infty}\int^{-N}_{-M}|f|+\int^M_N|f|\leq\varepsilon/2<\varepsilon$$ This shows that condition in (3) indeed yields convergence of the proper integral $\int^\infty_{-\infty}|f|$.

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Proof of Succifiency of Cauchy condition for improper integrals: Suppose the Cauchy condition in (1) holds. Given $\varepsilon>0$, there is $a>0$ such that $M'>N'>a$ and $M>N>a$ \begin{align} \Big|\int^{-N'}_{-M'}f+\int^M_Nf\Big|&<\varepsilon \end{align}

Let $(a_n)$ and $(b_n)$ be arbitrary sequences of positive numbers such that $a_n,b_n\xrightarrow{n\rightarrow\infty}\infty$. We will show that A. $I_n:=\int^{b_n}_{-a_n}f$ is a Cauchy sequence $$I=\lim_n\int^0_{-b_n}f+\int^{a_n}_0f$$ B. The limit $I$ is dependent of sequences $a_n$ and $b_n$. This will imply that $\lim\limits_{M,N\rightarrow\infty}\int^M_{-N}f=I$.

Let $k>0$ such that $a_n,b_n>a$ for all $n>k$. Then $$|I_m-I_n|=\big|\int^{-\min(a_n,a_m)}_{-\max(a_n,a_m)}f +\int^{\max(b_n,b_m)}_{\min(b_n,b_m)}f\Big|<\varepsilon $$ for all $m,n>k$. This proves A.

Let $a'_n,b_n$ be another pair of sequences of positive numbers with $a'_n,b'_n\xrightarrow{n\rightarrow\infty}$ and let $J_n=\int^{b'_n}_{a'_n}f$. As both $I_n$ and $J_n$ are Cauchy sequences, they converge to say $I$ and $J$ respectively. Then $$|I-J|\leq |I-I_n|+|I_n-J_n|+|J_n-J|$$

Choose $K>0$ so that $a_n,a'_n,b_n,b'_n>a$ whenever $n>K$. The Cauchy condition implies that $$|I_n-J_n|=\Big|\int^{-\min(a'_n,a_n)}_{-\max(a'n,a_n)}f +\int^{\max(b'_n,b_n)}_{\min(b'_n,b_n)}f|<\varepsilon$$ Hence, by letting $n\rightarrow\infty$ we obtain that $$|I-J|\leq\varepsilon$$ that is $I=J$. This proves B.

I leave to the OP to show that $\int^\infty_{-\infty}f$ converges iff for any sequences $(a_n)$ and $(b_n)$ of positive numbers such that $a_n,b_n\xrightarrow{n\rightarrow\infty}\infty$, $I_n=\lim_n\int^{b_n}_{a_n}f$ exists and the limit is independent on the particular sequences $a_n$ and $b_n$.