The word "induced" does not always refer to "a subset inheriting some sort of property from a larger set it belongs to". It also refers to one kind of structure defining another kind of structure on the same set. Examples:
- norm induced by an inner product
- metric induced by a norm
- topology induced by a metric
All of these can refer to the same underlying set, e.g. $\mathbb{R}^n$.
In the context of your quote at the beginning, I would interpret "induced" as in
measure induced by Riemannian metric [by means of the volume form].
Here is a usage example from MathOverflow.
But if you want to induce the measure on $S^2$ coming from the ambient space $\mathbb{R}^3$, I can suggest some approaches:
Consider the Hausdorff measure $\mathcal H^2$ on $\mathbb{R}^3$. Its restriction to $S^2$ is the surface measure.
Given $A\subset S^2$, define the [upper] Minkowski content of $A$ as
$$m^*(A) = \limsup_{r\to 0} \frac{\mu(A_r)}{2r}\quad \text{where } A_r=\{x\in\mathbb{R}^3 : \operatorname{dist} (x,A)\le r\} $$
For nice sets this is the same as surface area; unfortunately for uglier sets countable additivity fails. Also, the $\limsup$ and $\liminf$ are different in general.
Fix the subadditivity issue in #2 by defining the [upper] packing measure of $A$ as
$$\inf\left\{ \sum_{j\in\mathbb{N}} m^*(A_j) : \bigcup_{j\in\mathbb{N}} A_j = A\right\}$$
This is an ugly definition, but it has countable sub-additivity built-in. Since the resulting measure is invariant under isometries of $S^2$, it is in fact the surface area.
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.
Best Answer
I think your point 4 is the most important one to start the discussion with. Every mathematical tool has a domain of application which must be considered. The world of deterministic integrals appearing in applied mathematics, physics, engineering, chemistry and the rest of the sciences does not require the ability to have a well-defined integral for pathological functions like the question mark function. Keep in mind the historical context in which the Lebesgue integral was introduced: mathematicians had started realizing that many results which were claimed to have been proven for all functions actually could be contradicted by constructing abstract functions with paradoxical quantities - which had no analogue in the world of science. Thus the purpose of the Lebesgue integral was not to perform new integrals of interest to the sciences, but rather to place the mathematical formalism on a solid foundation. The Lebesgue integral is an antidote to a crisis in the foundations of mathematics, a crisis which was not felt in any of the sciences even as it was upending mathematics at the turn of the 20th century. "If it isn't broken, don't fix it" would be a natural response applied mathematicians and scientists could apply to this situation.
To anyone learning the Lebesgue integral for the purposes of expanding the scope of scientific integration they can perform, I would caution them with the previous paragraph.
However, the power of the Lebesgue integral (and the apparatus of measure theory) lies in its ability to make rigorous mathematical statements that apply to very badly behaved functions - functions that are so badly behaved, they often need to be specifically constructed for this purpose, and have no analogue in "real life". These are functions that are so delicate, an arbitrarily small "tweak" will destroy all these paradoxical properties. (This can be made rigorous in many ways, one of which is the fact that bounded continuous functions are dense in $L^p[0,1]$ - so for any terrible function $f\in L^p[0,1]$ and any $\epsilon>0$ you can find a very nice function $g$ with $|f(x)-g(x)|<\epsilon$ for all $x\in[0,1]$.) In "real life", all measurements carry errors and as a corollary any property that is destroyed by arbitrarily small modifications is not one that can actually be measured!
Despite all this, there is one major application of Lebesgue integration and the measure-theoretic apparatus: stochastic calculus, where one attempts to integrate against stochastic processes like Brownian motion. Even though the sample paths of Brownian motion are continuous, they represent the most badly behaved class of continuous paths possible and require special treatment. While the theory is very well developed in the Brownian case (and many of the top hedge funds on Wall Street have made lots of money exploiting this) there are other stochastic processes whose analysis is much more difficult. (How difficult? Well, the core of the Yang-Mills million dollar problem boils down to finding a way to rigorously define a certain class of very complicated stochastic integrals and show that they have the "obvious" required properties.)