Due to the completely different nature of the Riemann and Lebesgue approaches, there is no direct proof of the equivalence you mention as far as I know.
To show that the class of absolutely HK-integrable functions is the class of Lebesgue integrable functions, you use the following characterisation:
If $V$ is a vector space of Lebesgue measurable functions, and if $I : V \to \mathbb{R}$ is a linear map satisfying the following properties:
- $C([a, b]) \subset V$;
- $f \in V$ implies $|f| \in V$;
- if $(f_n)_n \subset V$ satisfy the assumptions of the monotone convergence theorem, then $\lim_n f_n \in V$ and $I(\lim_n f_n) = \lim_n I(f_n)$
then $V = L^1([a,b])$ and $I$ is the Lebesgue integral.
Therefore, you need to show:
- Riemann integrable functions are HK integrable;
- HK satisfies the Monotone convergence theorem;
- HK integrable functions are measurable.
The second one is classical and can be found in any book on the HK integral. The measurabilit part can be shown by approximating the function by step functions given by the Riemann sums. The proof is a bit technical though (but not too much).
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
As Martin pointed out in a comment, this does not mean that there are examples to the contrary, only that no proof is yet at hand.