User:Addemf/sandbox/Introduction to Differentiation

Derivative of the Integral in Elementary Analysis
Here is a statement which is common in an elementary analysis setting, and is often used as a tool to prove the Fundamental Theorem of Calculus.

Theorem: Let $$ f:\Bbb R\to\Bbb R$$ be a function integrable on $$\Bbb R$$ and continuous at a point $$x\in\Bbb R$$. Define the area function


 * $$F(x) = \int_{-\infty}^x f(x)\ dx$$

Then F is differentiable at x and


 * $$F'(x)=f(x)$$

Proof: Consider the difference quotient


 * $$g(h)=\frac{\int_{-\infty}^{x+h}f(t)\ dt - \int_{-\infty}^x f(t)\ dt}{x+h-x} = \frac 1 h \int_x^{x+h} f(t)\ dt$$

So $$F'(x)=\lim_{h\to 0} g(h)$$, which we need to demonstrate exists. For this we let $$\varepsilon\in\Bbb R^+$$ and consider, initially for any arbitrary h,


 * $$\left|\frac 1 h \int_x^{x+h}f(t)\ dt - f(x)\right| = \left|\frac 1 h\int_x^{x+h}f(t) \ dt - \frac 1 h\int_{x}^{x+h} f(x) \ dt\right|$$

which we have done so that we may now push the two integrals together, which in turn is useful because then we can apply the integral triangle equality.


 * $$=\left|\frac 1 h \int_x^{x+h} (f(t)-f(x))\ dt \right| \le \frac 1 h\int_x^{x+h}|f(t)-f(x)|\ dt $$

Now with this, we can use the continuity of f at x. Let $$h\in\Bbb R$$ such that for all $$t\in(x-h,x+h)$$, we have $$|f(t)-f(x)|<\varepsilon$$. For this choice of h, our earlier calculations show that


 * $$\frac 1 h\int_x^{x+h} |f(t)-f(x)|\ dt\le \frac 1 h \int_x^{x+h} \varepsilon \ dt = \varepsilon $$

Derivative of the Integral for a General Measurable Function
The problem with the theorem above is the requirement that f is continuous. So many of the functions we're interested in are measurable and not even continuous at any point -- the famous Dirichlet function is always a good example.

The proof above cannot be repeated for a general measurable function, and it is clear where continuity became important. However, this also suggests a point of focus: Can we show that


 * $$\lim_{h\to 0^+}\frac 1 h \int_x^{x+h} |f(t) -f(x)|\ dt=0$$

for an arbitrary measurable function?

One strategy leaps immediately to mind: Every measurable function is approximately continuous!

However, there are two problems that arise immediately. One is that the relevant theorem was stated only on a compact domain, and here we have domain equal to $$\Bbb R$$.

For another thing, the theorem is just false for arbitrary measurable functions. It is not too hard to come up with an example function which goes to infinity at a point, and at that point the equation above will be false.

Therefore we should not try to prove the claim on all of $$\Bbb R$$. Instead we should show that the claim holds almost everywhere.

Every Integrable Function Is Approximately Simple
In this subsection we want to resolve the other problem: We cannot directly use the approximation of measurable functions by continuous functions, because the result we obtained for this was constrained to a compact domain. We will try to show that, for any integrable $$f:\Bbb R\to\Bbb R$$ and $$\varepsilon\in\Bbb R^+$$, there is a continuous function $$g:\Bbb R\to\Bbb R$$ such that


 * $$\int_{\Bbb R}|f-g| < \varepsilon$$

Going directly toward that goal doesn't have an immediately clear path. Instead, we set a simpler goal, literally. We start by showing the corresponding result for simple functions, rather than continuous functions.

That is to say, here we will show that under the same conditions, there is a simple $$\psi:\Bbb R\to\Bbb R$$ such that


 * $$\int_{\Bbb R}|f-\psi|<\varepsilon$$

Because we assume f is integral as a general measurable function, we need to consider two integrals, $$\int_{\Bbb R}f^+$$ and $$\int_{\Bbb R}f^-$$. We know that each of these is finite and nonnegative.

By their definitions as nonnegative integrals, there exist simple functions such that $$0\le\varphi\le f^+$$ and $$0\le\psi\le f^-$$ and


 * $$\int_{\Bbb R}(f^+-\varphi), \quad \int_{\Bbb R}(f^--\psi) \quad < \varepsilon $$

Exercise 1. Approximate Simple Integration
Finish the proof started above. In particular, show that there is a simple function $$\chi$$, such that $$\int_{\Bbb R}|f-\chi|$$ is arbitrarily small.

(I am being deliberately coy about what the function $$\chi$$ is. It must be somehow related to $$\varphi,\psi$$ but I want you to think about exactly how.)

Integrable is Approximately Step
''Exercise 3. Approximate Step Integration'' is the focus of this subsection. However, we will need the result of ''Exercise 2. Restricting the Domain Makes Integrals Small will be helpful for us to complete Exercise 3.''

Exercise 2. Restricting the Domain Makes Integrals Small
Let $$f:\Bbb R\to\Bbb R$$ be an integrable function and $$\varepsilon\in\Bbb R^+$$ arbitrary.

Show that there exists a $$\delta\in\Bbb R^+$$ such that for every measurable subset $$E\subseteq\Bbb R$$ with $$ \lambda(E)<\delta$$ we have


 * $$\int_Ef < \varepsilon$$

Hint: Use $$\varepsilon$$ and the monotone convergence theorem to find a simple function $$0\le\psi\le f$$ such that $$\int f < \varepsilon + \int \psi $$.

Now select $$\delta$$ small enough to make $$\psi\delta <\varepsilon$$.

Use these ingredients to split $$\int f$$ into $$\int(g-\psi) + \int\psi$$ and proceed with the properties of integrals.

Exercise 3. Approximate Step Integration
Use the earlier result relating simple to step functions, to show that there is a step function s such that


 * $$\int|f-s|<\varepsilon$$

with f and $$\varepsilon$$ as before.

Hint: Split the integral into two domains, one of which has a small domain and the other has a small integrand. Show that both terms are small.

Exercise 4. Approximate Continuous Integration
With the assumptions as before, now show that there is a continuous function g such that $$\int_{\Bbb R}|f-g|$$ is small. Given what we've shown above, at some point in your solution you should observe that for any simple function $$\psi$$,


 * $$\int|f-g| \le \int_{\Bbb R}|f-\psi|+\int|\psi-g|$$

and the left-hand term can be chosen to be small. After making this choice, you should be able to then make a choice of g which makes the second term small.

Markov and Hardy-Littlewood
You've come a long way on this journey! There have been several nontrivial theorems in this lesson, all in service of proving one result.

I have bad news, I'm afraid. Not only are we still not ready for the final result, but we are not even very close.

There are two further results which we will need, one of them is easy to prove (Markov's inequality) and one of them harder (Hardy-Littlewood). It may feel like a long walk, but let me provide a small amount of cheer-leading: These two results are each


 * useful toward our current goals,
 * interesting (although de gustibus non est disputandum), and
 * serve as essential tools for later measure theory and, especially in the case of Markov's inequality, probability theory.

Both of them, in different senses, place a bound on the size of the domain of an integrable function where the function is sufficiently large. Markov's inequality places a bound on this set, by a direct comparison of the function['s absolute value]. Hardy-Littlewood uses a comparison with the function's average value.

In both cases, the ability to bound the set of points where the function is large, the allows us to argue that the value of the integral on this set is small. Of course, a tool which allows us to show certain quantities tend toward zero is often useful in analysis.

The Plan
Because they are somewhat more thematically and emphatically different from the results of the current lesson, we will prove the statement and proof of Markov and Hardy-Littlewood in their own separate lesson.

The plan for this section is to:


 * Prove Markov and Hardy-Littlewood.
 * Use this to conclude the proof of length-measure differentiation.
 * Introduce the concept of variation and its relationship to integration and differentiation.
 * Introduce the concept of absolute continuity and its relationship to area functions.

With that we will conclude our analysis of how differentiation relates to length-measure.

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