Function limit and sequence limit equivalent?

Question

I found this problem in a textbook, is to say if the statement it's true or false:

If $f$ is positive and $\lim_{n}\int_{1}^n\small f(t)dt \ = A$, then $\int_{1}^\infty \small f(t)dt \space\space$exists and is equal to $A$.

My intuiton says it's true, but how does one prove its validity?

Answer 1

Assume $f $ locally integrable at $[1,+\infty) $.

if $f $ is positive then the sequence $(\int_1^nf) $ is increasing and convergent. So, it is bounded above.

$$\exists M\in \mathbb R \; : \; \forall n\in\mathbb N \;\; \int_1^nf\le M $$

$$\implies \forall x\in [1,\infty) \; \int_1^xf\le \int_1^{\lfloor x \rfloor +1}f\le M $$

$$\implies \int_1^{+\infty}f \text { converges } $$

Answer 2

The following is a fact about limits that one can verify straight from their definitions:

Let $g\colon [1,\infty) \to \Bbb R$ be a function, and define the "sequence of sampled values" $a_n = g(n)$. If $\lim_{x\to\infty} g(x)$ exists and equals $A$, then $\lim_{n\to\infty} a_n = \lim_{n\to\infty} g(n)$ also exists and equals $A$.

Your question follows from taking $g(x) = \int_1^x f(t)\,dt$.

Answer 3

We have $\displaystyle\int_{1}^{\infty}f(t)dt\geq\int_{1}^{n}f(t)dt$ for each $n$, so taking $n\rightarrow\infty$ we have $\displaystyle\int_{1}^{\infty}f(t)dt\geq A$.

Now given any sequence $(x_{k})$ such that $x_{k}\rightarrow\infty$, then for each $k$, we find an $n_{k}$ such that $n_{k}>x_{k}$, and we can choose $(n_{k})$ to be increasing, then $\displaystyle\int_{1}^{x_{k}}f(t)dt\leq\int_{1}^{n_{k}}f(t)dt$, taking $k\rightarrow\infty$, we have $\displaystyle\int_{1}^{\infty}f(t)dt\leq A$.

Note that $\left(\displaystyle\int_{1}^{n_{k}}f(t)dt\right)_{k}$ is a subsequence of $\left(\displaystyle\int_{1}^{n}f(t)dt\right)_{n}$ and $\displaystyle\int_{1}^{\infty}f(t)dt$ is defined by $\lim_{x\rightarrow\infty}\displaystyle\int_{1}^{x}f(t)dt$, to realize the latter limit to sequential characterization, we need arbitrary sequence $(x_{k})$ such that $x_{k}\rightarrow\infty$ along with $\displaystyle\int_{1}^{x_{k}}f(t)dt$.

Actually one realizes to $x_{k}=k$, then we are done.

Thought:

Thing is not so easy if $f$ is not nonnegative and if the improper integral is not known whether it exists.

For example, given that $f\in L^{1}({\bf{R}})$, we are to ask whether the existence of
\begin{align*} \lim_{k\rightarrow\infty}\int_{|\xi|<2^{k}}\widehat{f}(\xi)e^{-2\pi ix\xi}d\xi=f(x) \end{align*} entails the existence of \begin{align*} \lim_{k\rightarrow\infty}\int_{|\xi|<k}\widehat{f}(\xi)e^{-2\pi ix\xi}d\xi=f(x), \end{align*} the answer is no, but both convergence are true, the first one can be proved by Littlewood-Paley Theorem while the latter is proved by Carleson-Hunt Theorem, which is extreme difficult.