Could we ever hope to integrate all functions? The Riemann integral has a weakness, in that it cannot integrate many functions of interest, such as Dirichlet's function $\boldsymbol{1}_\mathbb{Q}$. The Lebesgue and Henstock-Kurzweil integrals extend the range of functions which can be integrated, but still, there are functions which these integrals cannot deal with.
My question is thus. Let $\mathcal{A} \subseteq \mathcal{P}(\mathbb{R})$ be a "rich enough" family of "nice enough" subsets of $\mathbb{R}$ (e.g. $\mathcal{A}$ could be the collection of all (bounded?) Borel subsets of $\mathbb{R}$). Let $\overline{\mathbb{R}} = \mathbb{R} \cup \{ -\infty, +\infty \}$ be the extended real line.
Is it possible to define an "integration functional" $\mathbf{I}: \mathbb{R}^\mathbb{R} \times \mathcal{A} \to \overline{\mathbb{R}}$ satisfying the following:


*

*$\mathbf{I}$ is total (i.e. it gives a value for every function $f: \mathbb{R} \to \mathbb{R}$ and set $B \in \mathcal{A}$);

*$\mathbf{I}$ has the familiar "nice" properties we expect of integrals, e.g.:


*

*Linearity: $\mathbf{I}(cf+dg,B) = c\mathbf{I}(f,B) + d\mathbf{I}(g,B)$;

*Monotonicity: if $f(x) \leq g(x)$ for all $x \in B$, then $\mathbf{I}(f,B) \leq \mathbf{I}(g,B)$;

*Translation invariance: if $f_c$ is defined by $f_c(x) = f(x+c)$, then $\mathbf{I}(f,B) = \mathbf{I}(f_c,B+c)$;

*If $B,C \in \mathcal{A}$ are disjoint, then $\mathbf{I}(f,B \cup C) = \mathbf{I}(f,B) + \mathbf{I}(f,C)$;

*Perhaps there are other properties we might want that I've overlooked.


*Whenever $f: \mathbb{R} \to \mathbb{R}$ is (Riemann/Lebesgue/HK/etc)-integrable on $B$, we have $\mathbf{I}(f,B) = \int_B f$.


In other words, is it possible to define a notion of integration which extends standard integration, and can integrate any function? Ideally, we would want $\mathcal{A} = \mathcal{P}(\mathbb{R})$, but I recognise that we might have to restrict it to get a meaningful/nontrivial answer.
My bet is that the answer will be much the same as Vitali's theorem. I'm guessing that over $\mathsf{ZF}$, $\mathsf{AC}$ would prove there is no such $\mathbf{I}$, while other "nice" axioms such as $\mathsf{AD}$ might allow one to construct such an $\mathbf{I}$.
 A: Modulo the problem of assigning a value to "$\infty -\infty$" that can come up in the linearity property, such an object exists (assuming choice). [Note that this is a problem for the Lebesgue integral as well, so you can expect linearity only if at least one of the functions has a finite "integral"]
It is wellknown that there is a finitely (but not countably) additive measure $\mu:\mathcal P(\mathbb R)\rightarrow [0,\infty]$ extending the Lebesgue measure that is still translation invariant [cf. this Mathoverflow post]. The point is that one can construct such an $I$ that you are looking for from $\mu$ similar to how one consturcts the Lebesgue integral from the Lebegue measure. We must be careful that this goes through without $\sigma$-additivity.
I will make use of the suggestion of Hanuel Jeon and define a map $I:{}^{\mathbb R}\mathbb R\rightarrow \bar{\mathbb R}$ which satisfies the obvious modification of the properties you are looking for. One can translate $I$ to the setting in the post by defining $I^\prime(f, B)=I(f\cdot\chi_B)$.
As for the Lebesgue integral we start with simple functions. We call a function $f$ simple if it is of the form 
$$f=\sum_{i<n}\alpha_i\chi_{A_i}$$
Where $A_i\subseteq\mathbb R$ with $\mu(A_i)<\infty$ and $\alpha_i\geq 0$ for all $i<n$. For such $f$ we set $I(f)=\sum_{i<n}\alpha_i\mu(A_i)$. Now more generally for positive functions $g\geq 0$ we let 
$$I(g)=\operatorname{sup}\{I(f)\mid f\leq g\text{ is simple}\}$$
It is easy to check (using the monotonicity of $\mu$) that this definition agrees with the earlier one for simple functions $g$. 
It is clear that, up to now, $I$ is monotonic and translation invariant (as $\mu$ is). First lets check that $I$ agrees with the Lebesgue integral for a Lebesgue measurable positive function $g$. Clearly $\int g\ d\lambda\leq I(g)$ since any Lebegue measurable simple function is simple. To show $\geq$, let $f\leq g$ be simple. Write 
$$f=\sum_{i<n}\alpha_i\chi_{A_i}$$
Now we have 
$$I(f)=\sum_{i<n} \alpha_i\mu(A_i)\leq\sum_{i<n}\alpha_i\mu(\{g\geq\alpha_i\})=\sum_{i<n}\alpha_i\lambda(\{g\geq\alpha_i\})\leq\int g\ d\lambda$$
which is what we needed to show.
Linearity holds true as well, but is a little awkward to show. It is easy to see that $I(c g)=cI(g)$ for any positive $g$ and $c\geq 0$. Suppose we have two positive $g, h$. We have $I(g)+I(h)\leq I(g+h)$ since for any $g_0\leq g, h_0\leq h$ simple, $g_0+h_0\leq g+h$ is simple and $I(g_0+h_0)= I(g_0)+I(h_0)$ (which can be easily checked). To show $\geq$, given $f\leq g+h$ simple and $\epsilon>0$, we will produce simple $g_0\leq g$, $h_0\leq h$ such that $f\leq g_0+h_0+\epsilon$. (This suffices). Let $\alpha_\ast=\operatorname{max}(f)$
There is then a simple $g_0\leq g$ such that


*

*$g(x)-g_0(x)<\epsilon$ if $g(x)<\alpha_\ast$

*$g_0(x)=\alpha_\ast$ if $g(x)\geq\alpha_\ast$
Such a $g_0$ is not difficult to construct. If we let $h_0=\operatorname{max}\{f-g_0-\epsilon, 0\}$ then $h_0$ is simple and $h_0\leq h$. Thus $g_0, h_0$ are as desired. 
Finally we can define $I(f)=I(f^+)-I(f^-)$ for a general $f$ if at least one of $I(f^+), I(f^-)$ is finite, where $f^+=\operatorname{max}\{f, 0\}$ and $f^-=\operatorname{max}\{-f, 0\}$. All the nice properties we care about are preserved in this step.
$I$ has many strange properties. For example there is an everywhere positive function $g$ with $I(g)=0$. Furthermore, for the Lebegue integral it is the case that if $g$ is Lebesgue measurable and $(f_n)_{n\in\mathbb N}$ is increasing and converges to $g$ pointwise and all $f_n$ are Lebesgue simple, then $\int g\ d\lambda=\operatorname{sup}_{n\in\mathbb N} I(f_n)$. This is no longer true in our setting: $I(g)$ can be infinite while there is an increasing sequence $(f_n)_{n\in\mathbb N}$ of simple functions converging to $g$ pointwise with $I(f_n)=0$ for all $n$. Luckily we did not need this.
A final remark: You mentioned AD in the post. Under AD, all sets of reals and thus all functions $f:\mathbb R\rightarrow\mathbb R$ are Lebesgue measurable. Hence the Lebesgue integral itself is such an example.
Edit: As requested, here are some examples of functions with these "strange properties" described above. Using the Vitali set construction, we can find a partition $\langle A_i\mid i\in\mathbb N\rangle$ of $[0, 1]$ so that the $A_i's$ are translations of each other modulo the equivalence relation $x\cong y\Leftrightarrow x-y\in\mathbb Z$. Since $\mu$ is translation invariant, they must all have the same measure and as they partition a set of finite measure, they are of measure $0$. Now let $B_i=\bigcup_{j\in\mathbb Z} A_i+j$. Then the function 
$$g=\sum_{i=0}^{\infty} \frac{1}{i}\chi_{B_i}$$
is positive everywhere as the $B_i$ partition $\mathbb R$. However, if $A$ is a set with finite $\mu$-measure and $A\subseteq\{g\geq\alpha\}$ for some $\alpha>0$ then $\mu(A)=0$. If follows that for any simple $f\leq g$ we have $I(f)=0$ and thus $I(g)=0$.
For the second strange property, take the $A_i$ from above again and a bijection $h:\mathbb N\rightarrow\mathbb N\times\mathbb N$. Let
$$f_n=\sum_{i<n}\chi_{A_{h(i)_0}+h(i)_1}$$
Then all $f_n$ are simple with $I(f_n)=0$ and the sequence $(f_n)_{n\in\mathbb N}$ is increasing and converges pointwise to $g$ the constant function with value 1.
