I've got a continuous function on a set $A \subset \mathbb{R}^n$ and I want to approximate it's Lebesgue Integral by a sum.
I would consider a disjoint partition $(A_{i})_{i\in \mathbb{N}}$ of $A$ and claim that for any $\epsilon>0$ there exist a $N\in \mathbb{N}$ such that for any $n\geq N$
$$ \left|\sum_{i=1}^n \phi(v_{i}) \lambda(A_{i}) – \int_{A} \phi(v) \: d\lambda\right|<\epsilon $$
holds where $\phi(v_{i})$ is some value of $\phi$ on $A_{i}$.
Is this possibe?
Best Answer
Your method of computing the integral is exactly the definition of the Riemann integral. It is also known that the Lebesgue integral coincides with the Riemann integral when both are defined, so this is correct.
That answer was indeed terse, thanks @hardmath for bringing it up!
Edit
Your method of evaluating the integral incorporates the definition of the Riemann integral which is founded on partitions of the set over which the integral is being evaluated. It is also known that the Lebesgue integral coincides with the Riemann integral when both integrals are defined, so I think this approach is possible (with some caveats).
For instance in $\mathbb{R}$, we approximate the Riemann integral of $f(x)$ over some interval $(a, b) \subseteq \mathbb{R}$ as follows. First, we define a partition of $(a, b)$ into $k$ sub-intervals of equal length. Then, we choose arbitrary points $x_{1}$, $x_{2}$, $\dots$, $x_{k}$ from each of these sub-intervals. Then, we take the images of these points under $f$ to get $f(x_{1})$, $f(x_{2})$, $\dots$, $f(x_{k})$.
Each sub-interval has length $(b - a)/k$. From here, we approximate the integral by summing the areas of the rectangles formed over these sub-intervals of heights $f(x_{1})$, $f(x_{2})$, $\dots$, $f(x_{n})$ to get
$$\sum_{i = 0}^{k} f(x_{i}) \cdot \frac{b - a}{k}$$
This construction is very similar to the finite sum in OP's attempt; each sub-interval has Lebesgue measure $(b - a)/k$. Taking $n \to \infty$, we have the Riemann integral of $f(x)$ over $(a, b)$ so it is not surprising that the Riemann integral will coincide with the Lebesgue integral.
Without loss of generality, for intuition purposes, we assumed that $f$ is positive over the interval $(a, b)$. The signed integrals can be built on top of this naive definition.
In this multi-variable setting, the Riemann integral is generalized to $\mathbb{R}^n$ by using the same method over partitions into boxes of higher dimensions.
For instance, in $\mathbb{R}^3$, we partition the volume under surfaces into cuboids. In rigorous terms, these boxes of higher dimension are Cartesian products of intervals in $\mathbb{R}$. Also, the Lebesgue integral of each of these cuboids coincide with the volume classically defined as length $\times$ width $\times$ height.
In this case, for some surface $f(x, y)$, our sum would look like
$$\sum_{i \in I} f(P_{i}) \cdot V(A_{i})$$
for some index set $I$ that enumerates the squares $A_{i}$ that form the partition. Here, the $P_{i}$'s are the points chosen from each square $A_{i}$.
In the OP's attempt, we are using the Lebesgue integral, so our $A_{i}$'s do not necessarily need to be boxes in $\mathbb{R}^n$. In the Riemann integral, we are only limited to boxes (or countable unions of these boxes to form irregular shapes, see Peano-Jordan measure). But this limitation does not apply to the Lebesgue integral.
Hence, it is indeed possible, with the pre-requisite that the Lebesgue measure is defined on each of these $A_{i}$'s that make up the partition of the region of integration. Hence, the only caveat is that we require $A_{i}$'s to be Lebesgue measurable.