If $f$ is non negative and improper Riemann integrable on $[0, \infty)$, prove that $f$ is Lebesgue integrable and $\int_0^{\infty}f\mathrm{d}u = \lim_{n\rightarrow \infty} \int_0^nf\mathrm{d}u$. I need the link where I can get more information about this question and about improper Lebesgue integral in general.
[Math] Improper Lebesgue integral for non negative function
measure-theory
Related Solutions
Okay, after writing an answer for a long time I retract my comment: I don't think it is possible to give a canonical (in a sense which I'll explain soon) useful meaning to an "improper integral" in a general setting. I'll leave the point where the my answer broke as a reference (the previous answer can be seen in the end of the post).
Firstly, one important observation (in everything that follows, "integrable" means with respect to the Lebesgue sense): if $f: \mathbb{R}^n \to \mathbb{R}$ is integrable, then it is improperly integrable and to the same value (also, it doesn't depend on how you go to infinity). More precisely, let $A_i$ be any increasing sequence of sets such that $\bigcup A_i=\mathbb{R}^n$. Then - if $f$ is integrable - we have $$\lim \int_{A_i}f=\int_{\mathbb{R}^n}f .$$ This is a direct consequence of the dominated convergence theorem. The problem is when $f$ is not integrable (this is exactly like the contrast absolutely convergent/conditionally convergent. Even more so as we shall see.)
The case in $\mathbb{R}$ already shows that there is a huge issue: the "way" you take the limit matters (c.f. Cauchy principal value). Therefore, what makes sense to define is the following: having chosen an increasing sequence of sets $A_n$ such that $\bigcup A_n=X$ on a measure space $X$, let $$\int_X^{\operatorname{imp}(A_n)}f:=\lim \int_{A_n}f.$$ This depends on the choice of $A_n$, as it should. It is in this sense that I said that the definition is not "canonical". It coincides with the Lebesgue integral of $f$ if it is integrable (again due to the dominated convergence theorem). Note however that the above definition cannot even define the improper integral $\int_{-\infty}^\infty f$ (in its usual definition $\int_a^\infty f+\int_{-\infty}^bf$) properly.
One important observation is that the way the sequence is conceived is important, since we are taking a limit. This is in strong relation to the way a series can conditionally converge and not absolutely converge (indeed, a series is just an integral on a countable set). Note however that there is the general concept of a summable family (c.f. here) which if inspected closely may resemble the definition I assembled below: in fact, it is precisely the definition below in the case where we take the discrete topology (the compact sets on the discrete topology are the finite sets) and the counting measure. The concept below seems to enjoy the same good property as the improper integral: it coincides (in $\mathbb{R}$, and I believe that also in any $\sigma$-compact space) with the integral if the function is integrable. However, it does not generalize the concept of improper integral as you wanted, since they don't coincide even in $\mathbb{R}$. It is "another" integral.
Consider a locally compact Hausdorff space $X$ and a regular measure $\mu$ on $X$. Considering $\leq$ the inclusion on the set $\mathcal{K}$ of compact subsets of $X$, we have that $\mathcal{K}$ is directed, since finite union of compact sets is compact. Hence, given a real function $f: X \to \mathbb{R}$, we have a net $\lambda: \mathcal{K} \to \mathbb{R}$ given by $$\lambda_K:=\int_K f \, d\mu.$$ Then, define $$\int_X^{\operatorname{imp}} f:=\lim \lambda. $$
This coincides with the improper Lebesgue integral in $\mathbb{R}$ (!!no!!): for example, Let $\int_{[0,\infty)}^{\operatorname{impR}}$ denote the improper Lebesgue integral. Suppose $\int_{[0,\infty)}^{\operatorname{impR}} f=L$. Let $\varepsilon>0$. There exists $A>0$ such that if $x>A$ then $|L-\int_{[0,x)} f|<\varepsilon/4$. Note that this implies that $\left|\int_{[x,y]} f\right| <\varepsilon/2$ for every $x,y>A$, since $$\left|\int_{[x,y]} f\right| = \left|\int_{[0,y]}f-\int_{[0,x]}f\right| = \left|L-\int_{[0,x]} f + \int_{[0,y]} f-L\right| < \varepsilon/2. $$ Take $K=[0,A+1]$. Given any $K'>K$, then $K' \subset [0,B]$ for some $B>A+1.$ It follows that $$\left|L-\int_{K'} f\right|=\left|L-\int_Kf+\int_K f-\int_{K'} f\right| \leq \varepsilon/2+\left|\int_{K'-K}f\right|$$ Then, if $K'> K$.... the argument broke here. This won't work. Just take a non-integrable function and consider $K'$ a compact set big enough adjoining only portions where $f$ is positive. So, the notions don't coincide necessarily. However, it coincides if $f$ is integrable.
Since $f$ is nonnegative, the Lebesgue integral must exist (but may be infinite). With the existence of the improper Riemann integral we can show that the Lebesgue integral is finite and $f$ is "Lebesgue integrable" on $[a,b]$.
For all sufficiently large $n$ we have $[a+1/n,b] \subset [a,b]$. Since $f \chi_{[a+1/n,b]} \nearrow f$ as $n \to \infty$, it follows by the monotone convergence theorem that
$$\int_{[a,b]} f = \int_{[a,b]} \lim_{n \to \infty}f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{[a,b]} f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{a+ 1/n}^b f(x) \, dx = \int_a^bf(x) \, dx < +\infty$$
Here we have applied the equivalence of the Lebesgue and Riemann integrals on the interval $[a+1/n,b]$
Best Answer
I don't think there's a definition of "improper Lebesgue integrable": we have to make a difference between Riemman integrable functions and improper Riemman integrable functions since the latter is defined as a limit of integrals and not exactly as an integral. However, to solve the problem, you can use the Monotone Convergence Theorem. Given a subset $E$ of $\mathbb{R}$ let $\chi_E:\mathbb{R}\to \mathbb{R}$ be the indicator function of the set $E$, that is, $$ \chi_E(x)= \begin{cases} 1 & \mathrm{if}\, x\in E\\ 0 & \mathrm{if} \, x\not\in E\\ \end{cases} $$ Now, note that by definition $\int_0^n f \, dx= \int f\chi_{[0,n]}\, dx$. Define the sequence of functions $(f_n)$ given by $f_n=f\chi_{[0,n]}$ and note that $f_n\to f$ as $n\to \infty$ and that for every $n$, $f_n\leq f_{n+1}$. Since $f$ is nonnegative, all the $f_n$ are nonnegative. Also, since $f$ is improper Riemman integrable, we should have that $f$ is Riemman integrable on each of the intervals $[0, n]$ and since Riemman integrability implies Lebesgue integrability , we have that all of the $f_n$ are Lebesgue integrable. Hence, by the monotone convergence theorem, $$ \lim\limits_{n\to\infty}\int f_n\, dx= \int f\, dx $$ Since $\int f_n\, dx= \int_0^n f\, dx$ and the improper Riemman integral of $f$ is defined to be $\lim\limits_{n\to\infty}\int_0^n f\, dx$, then $$ \int f\, dx= \lim\limits_{n\to\infty}\int_0^n f\, dx<\infty $$ And by definition, this means that $f$ is Lebesgue integrable.