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.
- 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 $M,N\geq M_{0}.$
- 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.
- $f$ 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.
Best Answer
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.
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 $\int^\infty_{-\infty}f$.
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 $\int^\infty_{-\infty}|f|$.
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 $b_n$.