Measure Theory with respect to Soap Bubbles - 2

Since last time, we now know the crux of Measure Theory. So, let us proceed to our minimal surface. Its definition is derived quite a bit from its name: a surface with mean curvature as zero and which locally minimizes the area. Minimal surfaces solve the problem of minimizing the surface area in each class of surfaces. For example, in physical problems, a soap film spanning a wire frame naturally forms a minimal surface, as it minimizes the surface area for the given boundary.

We went over something called 'countable additivity' and though its implication remains the same, our application of it changes from a coordinate plane to a sphere (a soap bubble). The advantage of a sphere being Lebesgue measurable is that it can be broken down into smaller regions or sets such that countable additivity holds. One such way to think about it is to decompose the surface into two disjoint hemispheres. 

Moreover, this can be applied to multiple connected spheres or in our case, bubbles, to compute their surface area. While soap bubbles occur in Euclidean 3-dimensions, the surface is in 2-dimesnions and hence, the Lebesgue measure of the surface of the bubble is zero (a set of measure zero is measurable). Since we are interested in the surface of the sphere and to make things simpler, we will take its 2-D projection, i.e., a circle.

Say we divided the circle (surface of the sphere) in four equal disjoint sections as mentioned above. Their individual areas will be (4πr^2)/4 = πr^2. Finding the total surface area from hereon would not be tedious at all, perhaps trivial. Even so, to justify the triviality and the property, the TSA would follow from-

                                    S1+S2+S3+S4 = πr^2+πr^2+πr^2+πr^2=SA = 4πr^2

We are somewhere in the middle of theoretical understandings and practical applications, hence, there are many things to keep in mind, such as to check whether the domain is 'measurable'. If we want to be able to use the concept of Lebesgue Integral to compute the surface area of the sphere, we need our area function to be measurable and for this to hold good, we want our sets/intervals to be measurable and for  f(θ,∅)>a (follows from the definition of measurable functions).

Our area function in spherical coordinates is – SA = f(θ,∅) = r^2 sin⁡θ, which is continuous and bounded. 

f(θ,∅)>a, i.e., r^2 sin⁡θ>a, and so, sin⁡θ>a/r^2 

Now, we need our threshold value a to be bounded for the function to be greater than it. 

For θ varies from 0 to π, sin⁡θ varies from 0 to 1. So, a must be in the range 0<a<r^2. We were able to find a bound for a so, f(θ,∅)>a for all a. 

For the values of theta, we have interval [0,π]. For our azimuthal angle, ϕ, our interval is [0,2π]. By the definition of outer measure, we have, A = [0,2π] and we are taking E = [0,π]. Before we proceed, I will have to tell you to brace yourselves for some rigorous math. 

A∩E=m* (A∩[0,π])+m* (A∩[0,π]' ), here [0,π]'= (-∞,0)∪(π,∞)

A∩[0,π]=[0,π]

m* (A∩[0,π])=m* [0,π]=π

A∩[0,π]'=[0,2π]∩((-∞,0)∪(π,∞))=(π,2π]

m* (A∩[0,π]' )=m* (π,2π]=π

By adding we have,

RHS = m* (A∩E)+m*(A∩E')=π+π=2π and

LHS = m*(A)=m* ([0,2π])=2π       

(.∙. the outer measure is the length of the interval)

Hence, LHS = RHS and the inequality holds and our interval [0,2π] is measurable. Similarly, for when A = [0, π] and E = [0,2π], we have -

A∩E=m* (A∩[0,2π])+m* (A∩[0,2π]')

here [0,2π]'= (-∞,0)∪(2π,∞)

A∩[0,2π]=[0,π]

m*(A∩[0,2π])=m*[0,π]=π

A∩[0,2π]'=[0,π]∩((-∞,0)∪(2π,∞))=ϕ

m*(A∩[0,2π]')=ϕ

By adding we get,

RHS = m*(A∩E)+m*(A∩E' )=π+ϕ=π and

LHS = m*(A)=m*([0,π])=π           

Hence, LHS = RHS and the inequality holds and our interval [0,π] is measurable.

all in all, for those who skipped the above jargon, we showed our domain of these two intervals is measurable so, we can confirm that our function is also measurable and therefore, can be Lebesgue integrated. 

The interesting observation comes in when we calculate the upper and lower Lebesgue integrals and to save you from excessive steps, I have the two values: U(f)=4πr^2 and L(f)=0. 

If we recall, the Lebesgue integral is the common value between the two integral values however here that is not the case, at least in the traditional sense. But we also know Lebesgue integration is possible for our function so to keep our brain from going in circles, we can think of L(f) and U(f) as upper and lower bounds for our resulting Lebesgue Integral. We can see that L(f)≤4πr^2≤U(f). So, our Lebesgue Integral does exist.

We can be brave here and say that our attempt to connect a simple form of soap films in the 3-D Euclidean space with Measure theory has been successful. 

References:

1. He, S. (2014). Minimal Surfaces. https://vrs.amsi.org.au/wp-content/uploads/sites/6/2014/09/Shuhui_He_report_final.pdf 
2. Calegari, D. (2019). CHAPTER 3: MINIMAL SURFACES. In Minimal Surfaces. https://math.uchicago.edu/~dannyc/courses/minimal_surfaces_2014/minimal_surfaces_notes.pdf
3. Douglas, J. (1931). Solution of the Problem of Plateau. Transactions of the American Mathematical Society, 33(1), 263. https://doi.org/10.2307/1989472
4. Rad, T. (1930). The problem of the least area and the problem of Plateau. Mathematische Zeitschrift, 32(1), 763–796. https://doi.org/10.1007/bf01194665
5. Greene, R. E. (1993). Proceedings of symposia in Pure Mathematics (Vol. 54). American Mathematical Society.