Does integrability of real functions depend on the metric chosen? Although the Riemann integrability depends on limits, the Darboux integrability depends on the order defined (a function is Darboux integrable if the infimum of all the upper sums is equal to the supremum of all lower sums), since it uses the notion of supremum and infimum. So my question is - if I change the metric defined on the reals, will all standard integrable function will remain integrable with the same values?
EDIT: Suppose that we change the metric of both the domain and the range. Also, I am asking generally if Riemann integrable can become non-integrable and vice versa. I am talking about finite-interval integrability. Also, a general metric change with no special care about the topology.
First of all, let me spell out what I mean by Riemann integral with respect to the metrics $d$ on the range and $\rho$ on the domain of a function, restricting to bounded functions $f$ defined on closed intervals.
Let $P=(x_0<...<x_n)$ denote a partition of $[a,b]$ of mesh $m_\rho(P)$ (defined via the metric $\rho$). For this partition and a choice of sample points $t_i\in [x_{i-1}, x_{i}], t=(t_1,...,t_n)$, define the partial sum $$ S(f, P, t):= \sum_{i} d(f(t_i), 0) sign(f(t_i)) \rho(x_i, x_{i+1}). $$
Here $sign(y)= y/|y|$ if $y\ne 0$ and $sign(y)=0$ if $y=0$. Thus, this sum is the sum of "signed areas" of rectangles where the area is defined via the product metric on ${\mathbb R}^2$ defined via the metrics $\rho$ and $d$. Unlike what Alex M. is proposing, this definition is consistent with the more general notions of integration on metric / measure spaces. Riemann himself did not intend to define his integral beyond the realm of Euclidean spaces.
Then the Riemann integral $$ \int_{d,\rho, [a,b]} f(x)dx $$ is defined as the limit (if it exists) $$ \lim_{m_\rho(P)\to 0} S(f, P, t). $$
Now, consider some examples: Take the constant function $f(x)=1$ on $[0,1]$. Then $\int_0^1 f(x)dx=1$. But if you change the metric on $[0,1]$ by multiplying it by $a>0$ then the integral of the same function becomes $a$. Hence, even in the simplest case, you cannot hope to retain the values of integrals. Next, let me restrict to bounded functions on closed intervals. Then a function $f$ on $[a,b]$ is Riemann-integrable if and only if it is continuous almost everywhere in $[a,b]$ in the Lebesgue sense. Now, if $h: [a,b]\to [a,b]$ is a homeomorphism (a continuous monotonic function) which carries a Cantor set $C$ of zero Lebesgue measure to a Cantor set of positive Lebesgue measure then $\chi_C$ is Riemann-integrable but not $\chi_C \circ h^{-1}$. Hence, by changing the metric on $[a,b]$ via the function $h$: $$ \rho(x,y):= |h^{-1}(x)- h^{-1}(y)|, $$ does not preserve Riemann-integrability.
If you change the metric on the range while preserving the topology on the range then the notion of continuity does not change, hence, Riemann-integrability is preserved. But changing the metric on the target to, say, the discrete metric, makes every nonconstant continuous function on $[0,1]$ a discontinuous function. Therefore, say, the function $f(x)=x$ becomes discontinuous everywhere and hence, non-integrable.
The bottom line is that you want to retain the topology on the range and make an absolutely continuous change of topology on the domain. Then Riemann integrability is retained (for bounded functions).