Real world application of Lebesgue measure as opposed to Jordan measure

The answer to this is weird: you will never need to differentiate between Jordan measure and Lebesgue measure when computationally solving a specific real-world problem, but to coherently understand the way the world works you fundamentally and unavoidably need Lebesgue measure.

Real-world applications that use Lebesgue measure very rarely use it visibly. Since a bounded set is Jordan measurable whenever its boundary has measure $0$, Lebesgue measure is only necessary to measure extremely pathological sets, and the sorts of functions that are Lebesgue measurable but not Jordan measurable do not often come up in real-world applications. Furthermore, since computers can only make finitely many computations, the difference between finite additivity and countable additivity is immaterial to a computer. So, from a computational perspective, the concepts are the same.

However, this does not imply that Lebesgue measure can be replaced by Jordan measure when thinking about the real world. Tracing the reason why is a bit tedious and at first very theoretical, but worthwhile. Lebesgue measure has a number of important theoretical features that Jordan measure does not. For example, the space of Jordan $p$-integrable functions on $[0,1]$ (or $\mathbb{R}$, or most spaces) is not complete under any of the $p$-norms defined by $$||f||_p = \left(\int_{[0,1]} |f(x)|^pdx\right)^{1/p}.$$ To see this, consider the indicator of $[0,1] \cap \mathbb{Q}$, for instance.

The most important of these spaces from the point of applications is $L^2([0,1])$, the space of Lebesgue measurable functions with finite $2$-norm. The completeness of $L^2$ has deep ramifications. It is the key ingredient in making decomposition of functions into Fourier series work. Fourier series, in turn, play a central role in the physics behind modelling distributions of heat, waves, and a host of other important physical phenomena. Without the completeness of $L^2$, and hence without Lebesgue measure, none of this makes sense.

The completeness of $L^2$ also leads to more math that in turn finds wide applications. If you really, really trace the math, the completeness of $L^2$ is among a host of necessary facts that sit behind the complex measure case of the stunningly beautiful Riesz Representation Theorem. This theorem finds abundant applications in probability theory, and hence in the study of randomness in the real world. It is also key to the proof of the spectral theorem in operator theory. Operator theory is, in turn, the foundation that our modern understanding of quantum physics is built on.

Hence, our modern understanding of many physical processes, randomness, and quantum physics all rely deeply on mathematical features of Lebesgue measure that are not shared by Jordan measure. You don't clearly see these features when doing the end-stage computations to solve these problems, but they're unavoidably behind the relationships that allow for computations done with Jordan measure to work.

Not at all. For an example, get a continuous function $f$ defined on $[-\pi,\pi]$ such that $f(-\pi)=f(\pi)$ and perform the decomposition of $f$ in its fundamental harmonics, i.e. take its Fourier transform $\hat{f}$. Now ask yourself what happens if you synthetize symmetrically these harmonics, i.e. if you get $$\sum_{n=-N}^N \hat{f}(n)e^{int},$$ and then try to take the limit as $N\rightarrow\infty$. Will this series (known as Fourier series of $f$) converge pointwise to $f$? If not, is it possible to describe the set where this series does not converge pointwise to $f$? Well, you can guarantee that the set of points where such a series converges pointwise to $f$ has a complement of Lebesgue measure zero. In a sense, this is the best result you can obtain in general, because you can prove that for each set of Lebesgue measure zero, there exists a function $f$ of the type considered above whose Fourier series diverges at least on this set (an maybe at some other points).

So, coming back to earth, get the set $\mathcal{Q}$ of the rational points in $[-\pi,\pi]$. Then, since $\mathcal{Q}$ has Lebesgue measure $0$, there exists a continuous function $f$ defined on $[-\pi,\pi]$ such that $f(-\pi)=f(\pi)$, whose Fourier series diverges at least on $\mathcal{Q}$. However, we know that the set where the Fourier series of $f$ converges pointwise has a complement, say $\mathcal{P}$, whose Lebesgue measure is zero. Then the set where $f$ diverges, i.e. $\mathcal{P}$, cannot cointain any interval (otherwise the Lebesgue measure of $\mathcal{P}$ is greater than $0)$ and then has inner Jordan measure equal to $0$. On the other hand, $\mathcal{P}$ has outer Jordan measure equal to $2\pi$, because every interval that cointains $\mathcal{P}$ also cointains $\mathcal{Q}$, and the only interval that contains $\mathcal{Q}$ is $[-\pi,\pi]$. So $\mathcal{P}$ has different inner Jordan measure and outer Jordan measure and so it isn't Jordan measurable.

In conclusion:

There exists a continuous function defined on $[-\pi,\pi]$ that satisfies $f(-\pi)=f(\pi)$, whose Fourier series diverges precisely on a set that is Lebesgue measurable but not Jordan measurable.