4.1 KiB
Math4121 Lecture 38
Extended fundamental theorem of calculus with Lebesgue integration
Hardy-Littlewood maximal function
Given integrable $f$m and an interval I, look at the averaging operator A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(y)dy.
The maximal function is defined as
f^*(x)=\sup_{I \text{ is an open interval}} A_I f(x)
Theorem Hardy-Littlewood Maximal Function Theorem
Fix f integrable. For each \lambda>0, we define
E_\lambda^*=\{x\in\mathbb{R}: |f^*(x)|>\lambda\}
Then
m(E_\lambda^*)\leq \frac{2}{\lambda}\int_\mathbb{R} |f| dm
To give context for the maximal estimate, for any f integrable, \lambda>0,
E_\lambda=\{x\in\mathbb{R}: |f(x)|>\lambda\}
Then we have Marknov's inequality, m(E_\lambda)\leq \frac{1}{\lambda}\int_\mathbb{R} |f| dm. We know |f(x)|>\lambda \chi_{E_\lambda}(x) so Markov's inequality follows by integrating.
Proof:
Let f^*(x)=\sup_{I \text{ is an open interval such that } x\in I} \frac{1}{m(I)}\int_I f \, dm.
If x\in E_\lambda^*, then \exists I open interval such that x\in I and \frac{1}{m(I(x))}\left|\int_{I(x)} f \, dm\right|>\lambda.
Take K\subset E_\lambda^* compact. Then K\subset \bigcup_{x\in K} I(x). Taking the finite subcover, we have I_1, \ldots, I_n open intervals such that K\subset \bigcup_{i=1}^n I_i.
If three intervals, I,J,K have non-empty intersection, then one is contained in the union of the other two.
In particular, we can find another subcover for K, J_1, \ldots, J_N such that they have overlap of at most 2 (otherwise, we can remove the cover). We can state this as
\sum_{j=1}^N \chi_{J_j}(x)\leq 2
\begin{aligned}
m(K)&\leq \sum_{j=1}^N m(J_j)\\
&\leq \sum_{j=1}^N \frac{1}{\lambda}\left|\int_{J_j} f \, dm\right|\\
&\leq \frac{1}{\lambda}\int_\mathbb{R} \sum_{j=1}^N \chi_{J_j}(x) |f(x)| \, dx\\
&\leq \frac{2}{\lambda}\int_\mathbb{R} |f(x)| \, dx
\end{aligned}
Since A_I f(x) is measurable, f^* is measurable function and E_\lambda is measurable, we have
m(E_\lambda^*)\leq \frac{2}{\lambda}\int_\mathbb{R} |f(x)| \, dx
QED
3 Big Convergence Theorems
Theorem L.1 (Monotone Convergence Theorem)
Theorem L.2 (Fatou's Lemma)
Let \{f_n\}_{n=1}^\infty be a sequence of non-negative measurable functions on E. Then
\int_E \liminf_{n\to\infty} f_n \, dm\leq \liminf_{n\to\infty} \int_E f_n \, dm
Proof:
Let g_n=\inf_{k\geq n} f_k is a monotone increasing nonnegative, and the following properties holds:
\lim_{n\to\infty} g_n(x)=\sup_{n\geq 1} \inf_{k\geq n} f_k(x)=\liminf_{n\to\infty} f_n(x)
\int_E g_n \, dm\leq \inf_{k\geq n} \int_E f_k\, dm
So,
\int_E g_n \, dm\leq \inf_{k\geq n} \int_E f_k \, dm
Apply the monotone convergence theorem to g_n, we have
\lim_{n\to\infty} \int_E g_n \, dm\leq \liminf_{n\to\infty} \int_E f_n \, dm
QED
Theorem L.3 (Dominated Convergence Theorem)
Let \{f_n\}_{n=1}^\infty be a sequence of measurable functions on \mathbb{R} converging to f almost everywhere. If there exists integrable g such that |f_n|\leq |g| for all n, then
\int_E f \, dm=\lim_{n\to\infty} \int_E f_n \, dm
Proof:
Consider the function g+f_n and g-f_n, these are non-negative sequences of measurable functions. By Fatou's lemma, we have
\int g\,dm+\int f\,dm=\int_E \liminf_{n\to\infty} (g+f_n) \, dm\leq \liminf_{n\to\infty} \int_E (g+f_n) \, dm=\int_E g\,dm+\liminf_{n\to\infty} \int_E f_n\,dm
So, \int_E f\,dm\leq \liminf_{n\to\infty} \int_E f_n\,dm.
Similarly, we have
\int g\,dm-\int f\,dm=\int_E \liminf_{n\to\infty} (g-f_n) \, dm\leq \liminf_{n\to\infty} \int_E (g-f_n) \, dm=\int_E g\,dm-\limsup_{n\to\infty} \int_E f_n\,dm
So, \limsup_{n\to\infty} \int_E f_n\,dm\leq \int_E f\,dm.
So \limsup_{n\to\infty} \int_E f_n\,dm\leq \int_E f\,dm\leq \liminf_{n\to\infty} \int_E f_n\,dm.
Since \limsup_{n\to\infty} \int_E f_n\,dm\geq \liminf_{n\to\infty} \int_E f_n\,dm, we have \int_E f\,dm=\lim_{n\to\infty} \int_E f_n\,dm.
QED