## Thursday, June 15, 2017

### Perfect Competition With An Uncountable Infinity Of Firms

1.0 Introduction

Consider a partial equilibrium model in which:

• Consumers demand to buy a certain quantity of a commodity, given its price.
• Firms produce (supply) a certain quantity of that commodity, given its price.

This is a model of perfect competition, since the consumers and producers take the price as given. In this post, I try to present a model of the supply curve in which the managers of firms do not make systematic mistakes.

This post is almost purely exposition. The exposition is concrete, in the sense that it is specialized for the economic model. I expect that many will read this as still plenty abstract. (I wish I had a better understanding of mathematical notation in HTML.) Maybe I will update this post with illustrations of approximations to integrals.

2.0 Firms Indexed on the Unit Interval

Suppose each firm is named (indexed) by a real number on the (closed) unit interval. That is, the set of firms, X, producing the given commodity is:

X = (0, 1) = {x | x is real and 0 < x < 1}

Each firm produces a certain quantity, q, of the given quantity. I let the function, f, specify the quantity of the commodity that each firm produces. Formally, f is a function that maps the unit interval to the set of non-negative real numbers. So q is the quantity produced by the firm x, where:

q = f(x)
2.1 The Number of Firms

How many firms are there? An infinite number of decimal numbers exist between zero and unity. So, obviously, an infinite number of firms exist in this model.

But this is not sufficient to specify the number of firms. Mathematicians have defined an infinite number of different size infinities. The smallest infinity is called countable infinity. The set of natural numbers, {0, 1, 2, ...}; the set of integers, {..., -2, -1, 0, 1, 2, ...}; and the set of rational numbers can all be be put into a one-to-one correspondence. Each of these sets contain a countable infinity of elements.

But the number of firms in the above model is more than that. The firms can be put into a one-to-one correspondence with the set of real numbers. So there exist, in the model, a uncountable infinity of firms.

2.2 To Know

Cantor's diagonalization argument, power sets, cardinal numbers.

3.0 The Quantity Supplied

Consider a set of firms, E, producing the specified commodity, not necessarily all of the firms. Given the amount produced by each firm, one would like to be able to say what is the total quantity supplied by these firms. So I introduce a notation to designate this quantity. Suppose m(E, f) is the quantity supplied by the firms in E, given that each firm in (0, 1) produces the quantity defined by the function f.

So, given the quantity supplied by each firm (as specified by the function f) and a set of firms E, the aggregate quantity supplied by those firms is given by the function m. And, if that set of firms is all firms, as indexed by the interval (0, 1), the function m yields the total quantity supplied on the market.

Below, I consider for which set of firms m is defined, conditions that might be reasonable to impose on m, a condition that is necessary for perfect competition, and two realizations of m, only one of is correct.

You might think that m should obviously be:

m(E, f) = ∫Ef(x) dx

and that the total quantity supplied by all firms is:

Q = m((0,1), f) = ∫(0, 1) f(x) dx

Whether or not this answer is correct depends on what you mean by an integral. Most introductory calculus classes, I gather, teach the Riemann integral. And, with that definition, the answer is wrong. But it takes quite a while to explain why.

3.1 A Sigma Algebra

One would like the function m to be defined for all subsets of (0, 1) and for all functions mapping the unit interval to the set of non-negative real numbers. Consider a "nice" function f, in some hand-waving sense. Let m be defined for a set of subsets of (0, 1) in which the following conditions are met:

• The empty set is among the subsets of (0, 1) for which m is defined.
• m is defined for the interval (0, 1).
• Suppose m is defined for E, where E is a subset of (0, 1). Let Ec be those elements of (0, 1) which are not in E. Then m is defined for Ec.
• Suppose m is defined for E1 and E2, both being subsets of (0, 1). Then m is defined for the union of E1 and E2.
• Suppose m is defined for E1 and E2, both being subsets of (0, 1). Then m is defined for the intersection of E1 and E2.

One might extend the last two conditions to a countable infinity of subsets of (0, 1). As I understand it, any set of subsets of (0, 1) that satisfy these conditions is a σ-algebra. A mathematical question arises: can one define the function m for the set of all subsets of (0, 1)? At any rate, one would like to define m for a maximal set of subsets of (0, 1), in some sense. I think this idea has something to do with Borel sets.

3.2 A Measure

I now present some conditions on this function, m, that specifies the quantity supplied to the market by aggregating over sets of firms:

• No output is produced by the empty set of firms:
• m(∅, f) = 0.
• For any set of firms in the sigma algebra, market output is non-negative:
• m(E, f) ≥ 0.
• For disjoint sets of firms in the sigma algebra, the market output of the union of firms is the sum of market outputs:
• If E1E1 = ∅, then m(E1E1, f) = m(E1, f) + m(E2, f)

The last condition can be extended to a countable set of disjoint sets in the sigma algebra. With this extension, the function m is a measure. In other words, given firms indexed by the unit interval and a function specifying the quantity supplied by each firm, a function mapping from (certain) sets of firms to the total quantity supplied to a market by a set of firms is a measure, in this mathematical model.

One can specify a couple other conditions that seem reasonable to impose on this model of market supply. A set of firms indexed by an interval is a particularly simple set. And the aggregate quantity supplied to the market, when each of these firms produce the same amount is specified by the following condition:

Let I = (a, b) be an interval in (0, 1). Suppose for all x in I:

f(x) = c

Then the quantity supplied to the market by the firms in this interval, m(I, f), is (b - a)c.

3.3 Perfect Competition

Consider the following condition:

Let G be a set of firms in the sigma algebra. Define the function fG(x) to be f(x) when x is not an element of G and to be 1 + f(x) when x is in G. Suppose G has either a finite number of elements or a countable infinity number of elements. Then:

m((0,1), f) = m((0,1), fG)

One case of this condition would be when G is a singleton. The above condition implies that when the single firm increases its output by a single unit, the total market supply is unchanged.

Another case would be when G is the set of firms indexed by the rational numbers in the interval (0, 1). If all these firms increased their individual supplies, the total market supply would still be unchanged.

Suppose the demand price for a commodity depends on the total quantity supplied to the market. Then the demand price would be unaffected by both one firm changing its output and up to a countably infinite number of firms changing their output. In other words, the above condition is a formalization of perfect competition in this model.

4.0 The Riemann Integral: An Incorrect Answer

I now try to describe why the usual introductory presentation of an integral cannot be used for this model of perfect competition.

Consider a special case of the model above. Suppose f(x) is zero for all x. And suppose that G is the set of rational numbers in (0, 1). So fG is unity for all rational numbers in (0, 1) and zero otherwise. How could one define ∫(0, 1)fG(x) dx from a definition of the integral?

Define a partition, P, of (0, 1) to be a set {x0, x1, x2, ..., xn}, where:

0 = x0 < x1 < x2 < ... < xn = 1

The rational numbers are dense in the reals. This implies that, for any partition, each subinterval, [xi - 1, xi] contains a rational number. Likewise, each subinterval contains an irrational real number.

Define, for i = 1, 2, ..., n the two following quantities:

ui = supremum over [xi - 1, xi] of fG(x)

li = infimum over [xi - 1, xi] of fG(x)

For the function fG defined above, ui is always one, for all partitions and all subintervals. For this function, li is always zero.

A partition can be pictured as defining the bases of successive rectangles along the X axis. Each ui specifies the height of a rectangle that just includes the function whose integral is being sought. For a smooth function (not our example), a nice picture could be drawn. The sum of the areas of these rectangles is an upper bound on the desired integral. Each partition yields a possibly different upper bound. The Riemann upper sum is the sum of the rectangles, for a given partition:

U(fG, P) = (x1 - x0) u1 + ... + (xn - xn - 1) un

For the example, with a function that takes on unity for rational numbers, the Riemann upper sum is one for all partitions. The Riemann lower sum is the sum of another set of rectangles.

L(fG, P) = (x1 - x0) l1 + ... + (xn - xn - 1) ln

For the example, the Riemann lower sum is zero, whatever partition is taken.

The Riemann integral is defined in terms of the least upper bound and greatest lower bound on the integral, where the upper and lower bounds are given by Riemann upper and lower sums:

Definition: Suppose the infimum, over all partitions of (0, 1), of the set of Riemann upper sums is equal to the supremum, also over all partitions, of the set of Riemann lower sums. Let Q designate this common value. Then Q is the value of the Riemann integral:

Q = ∫(0, 1)fG(x) dx

If the infimum of Riemann upper sums is not equal to (exceeds) the supremum of the Riemann lower sums, then the Riemann integral of fG is not defined.

In the case of the example, the Riemann integral is not defined. One cannot use the Riemann integral to calculate the changed market supply from a countably infinite firms each increasing their output by one unit.

5.0 Lebesque Integration

The Riemann integral is based on partitioning the X axis. The Lebesque integral, on the other hand, is based on partitioning the Y axis, in some sense. Suppose one has some measure of the size of the set in the domain of a function where the function takes on some designated value. Then the contribution to the integral for that designated value can be seen as the product of that value and that size. The integral of a function can then be defined as the sum, over all possible values of the function, of such products.

5.1 Lebesque Outer Measure

Consider an interval, I = (a, b), in the real numbers. The (Lebesque) measure of that set is simply the length of the interval:

m*(I) = b - a

Let E be a set of real numbers. Let {In} be a set of an at most countable infinite number of open intervals such that

E is a subset of ∪ In

In other words, {In} is an open cover of E. The (Lebesque) measure of E is defined to be:

m*(E) = inf [m*(I1) + m*(I2) + ...]

where the infimum is taken over the set of countably infinite sets of intervals that cover E.

The Lebesque measure of any set that is at most countably infinite is zero. So the rational numbers is a set of Lebesque measure zero. So is a set containing a singleton.

A measurable set E can be used to decompose any other set A into those elements of that set that are also in E and those elements that are not. And the measure of A is the sum of the measures of those two set.

If a set is not measurable, there exists some set A where that sum does not hold. Given the axiom of choice non-measurable sets exist. As I understand it, the set of all measurable subsets of the real numbers is a sigma algebra.

5.2 Lebesque Integral for Simple Functions

Let E be a measurable subset of the real numbers. Define the characteristic function, χE(x), for E, to be one, if x is an element of E, and zero, if x is not an element of E.

Suppose the function g takes on a finite number of values {a1, a2, ..., an}. Such a function is called a simple function. Let Ai be the set of real numbers where gi = ai. The function g can be represented as:

g(x) = a1 χA1(x) + ... + an χAn(x)

The integral of such a simple function is:

g(x) dx = a1 m*(A1) + ... + an m*(An)

This definition can be extended to non-simple functions by another limiting process.

5.3 Lebesque Upper and Lower Sums and the Integral

The Lebesque upper sum of a function f is:

UL(E, f) = sup over simple functions gf of ∫Eg(x) dx

One function is greater than or equal to another function if the value of the first function is greater than or equal to the value of the second function for all points in the common domain of the functions. The Lebesque lower sum is:

LL(E, f) = inf over simple functions gf of ∫Eg(x) dx

Suppose the Lebesque upper and lower sums are equal for a function. Denote that common quantity by Q. Then this is the value of the Lebesque integral of the function.

Q = ∫Ef(x) dx

When the Riemann integral exists for a function, the Lebesque integral takes on the same value. The Lebesque integral exists for more functions, however. The statement of the fundamental theorem of calculus is more complicated for the Lebesque integral than it is for the Riemann integral. Royden (1968) introduces the concept of a function of bounded variation in this context.

5.4 The Quantity Supplied to the Market

So the quantity supplied to the market by the firms indexed by the set E, when each firm produces the quantity specified by the function f is:

m(E, f) = ∫Ef(x) dx

where the integral is the Lebesque integral. In the special case, where the firms indexed by the rational numbers in the interval (0, 1) each supply one more unit of the commodity, the total quantity supplied to the market is unchanged:

Q = ∫(0, 1)fG(x) dx = ∫(0, 1)f(x) dx

Here is a model of perfect competition, in which a countable infinity of firms can vary the quantity they produce and, yet, the total market supply is unchanged.

6.0 Conclusion

I am never sure about these sort of expositions. I suspect that most of those who have the patience to read through this have already seen this sort of thing. I learn something, probably, by setting them out.

I leave many questions above. In particular, I have not specified any process in which the above model of perfect competition is a limit of models with n firms. The above model certainly does not result from taking the limit at infinity of the number of firms in the Cournot model of systematically mistaken firms. That limit contains a countably infinite number of firms, each producing an infinitesimal quantity - a different model entirely.

I gather that economists have gone on from this sort of model. I think there are some models in which firms are indexed by the hyperreals. I do not know what theoretical problem inspired such models and have never studied non-standard analysis.

Another set of questions I have ignored arises in the philosophy of mathematics. I do not know how intuitionists would treat the multiplication of entities required to make sense of the above. Do considerations of computability apply, and, if so, how?

Some may be inclined to say that the above model has no empirical applicability to any possible actually existing market. The above mathematics is not specific to the economics model. It is very useful in understanding probability. For example, the probability density function for any continuous random variable is only defined up to a set of Lebesque measure zero. And probability theory is very useful empirically.

Appendix: Supremum and Infimum

I talk about the supremum and the infimum of a set above. These are sort of like the maximum and minimum of the set.

Let S be a subset of the real numbers. The supremum of S, written as sup S, is the least upper bound of S, if an upper bound exists. The infimum of S is written as inf S. It is the greatest lower bound of S, if a lower bound exists.

References
• Robert Aumann (1964). Markets with a continuum of traders. Econometrica, V. 32, No. 1-2: pp. 39-50.
• H. L. Royden (1968). Real Analysis, second edition.

Brian Romanchuk said...

The thing that bothers me: how can firms change hours worked (to optimise profits) without dragging in the household optimisation problem? Both optimisation problems will interact with each other, as well as government spending and the household budget constraint, which are specified as aggregate values.

Blissex said...

«For example, the probability density function for any continuous random variable is only defined up to a set of Lebesque measure zero.»

Ahem ahem, a number of sensible statisticians I suspect would have serious misgivings with any statement involving statistics and infinities or infinitesimals. It's a contentious issue.

«And probability theory is very useful empirically.»

Sure, but you are talking about a particular model/interpretation of probability theory, one that involves deep mathematical trickery that. A completely unrealistic model can still be an empirically useful model most of the time.
For example in signal processing it is often the case that a discrete signal is approximated with a continuous one, because the latter is more tractable symbolically, regardless of whether the continuous model is well-founded intellectually. "As long as it works". Physicists do a lot of the same misuse of cod-maths "because it works" "in most cases", without questioning the assumptions. As long as the bombs explode they get a lot of funding, very empirically :-).

This is a problem that mathematicians themselves have had for a long time: much mathematics empirically worked most of the time even if its "foundations" were broken. Because actually "weak logics" tend to empirically work most of the time too. But even many logicians ignore that the study and use of "weak logics" is a fruitful and interesting topic...

Blissex said...

«How can firms change hours worked (to optimise profits) without dragging in the household optimisation problem? Both optimisation problems will interact with each other,»

Wave your hands and say "the auctioneer", "tatonnement", "rational expectations" at least three times. It usually works :-).