To demonstrate that:

$\int_a^b \left(f(x) + g(x)\right) dx = \int_a^b f(x) dx + \int_a^b g(x) dx$

using Riemann sums, let's proceed step by step.

Step 1: Recall the definition of the Riemann sum.

The Riemann sum for a function $f(x)$ over the interval $[a, b]$ is given by:

$S(f) = \sum_{i=1}^n f(x_i^*) \Delta x$

where $\Delta x = \frac{b - a}{n}$ is the width of each subinterval, and $x_i^*$ is a point within the $i$-th subinterval $[x_{i-1}, x_i]$.

Step 2: Riemann sum for the sum of two functions.

Now, consider the Riemann sum for $f(x) + g(x)$:

$S(f + g) = \sum_{i=1}^n \left( f(x_i^*) + g(x_i^*) \right) \Delta x$

Step 3: Use properties of summation.

By the linearity of sums, we can split the sum:

$S(f + g) = \sum_{i=1}^n f(x_i^*) \Delta x + \sum_{i=1}^n g(x_i^*) \Delta x$

This is simply the sum of the Riemann sums for $f(x)$ and $g(x)$:

$S(f + g) = S(f) + S(g)$

Step 4: Take the limit as $n \to \infty$.

As $n \to \infty$, the Riemann sums converge to the definite integrals:

$\lim_{n \to \infty} S(f + g) = \int_a^b \left(f(x) + g(x)\right) dx$ and $\lim_{n \to \infty} S(f) = \int_a^b f(x) dx, \quad \lim_{n \to \infty} S(g) = \int_a^b g(x) dx$

Thus, we conclude that:

$\int_a^b \left(f(x) + g(x)\right) dx = \int_a^b f(x) dx + \int_a^b g(x) dx$

This completes the proof using Riemann sums.

Do you need further details on any part? Here are some related questions:

1. What are the key properties of Riemann sums used in this proof?
2. How can this result be extended to more than two functions?
3. Can the result be shown using different approaches, like limits?
4. What are the conditions for the existence of these integrals?
5. How do we interpret this result geometrically in terms of areas under curves?

Tip: Always ensure that the functions you are integrating are integrable (bounded and continuous on the interval) to apply the Riemann sum definition properly.