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)

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