Lebesgue Integral

Motivation §

The classical notion of the integral often builds from Riemann’s idea of an integral. Consider a function $f:[a,b]\to \mathbb{R}$. In Riemann’s process, the domain of the function is split into intervals of decreasing width, partitioning the domain evenly and infinitessimally. Rectangles (sometimes trapezoids) are constructed from each partition, with the height being the height of $f$ on that interval. The Riemann integral is then the limit of the sum of the areas of all these rectangles.

In general, however, the Riemann integral is too narrow. In particular, many classes of functions are discontinuous or otherwise unruly within the desired interval of integration. To extend the idea of integrals further, we look to a concept from measure theory: the Lebesgue integral.

Definition §

A measurable function $f:[a,b]\to \mathbb{R}$ is integrable if and only if for some sequence of continuous functions $\{f_n{\}}_{n=1}^{\infty }$ where $f_n:[a,b]\to \mathbb{R}$, the following conditions hold:

1. $f_n(x)\underset{n\to \infty }{\to }f(x)$ a.e.
2. ${\int }_a^b|f_n(x)-f_m(x)|dx\underset{n,m\to \infty }{\to }0$

When this is true, the sequence $I_n={\int }_a^b{f}_n(x)dx$ is Cauchy. The finite limit $I={\lim }_{n\to \infty }{I}_n$ is called the Lebesgue integral of $f$ on $[a,b]$ (also denoted $I={\int }_a^{b}f(x)dx$).

Over all the reals, $f:\mathbb{R}\to \mathbb{R}$ is integrable if and only if $\underset{a<b}{\sup }{\int }_a^b|f|={\lim }_{(a,b)\to (-\infty ,\infty )}{\int }_a^b|f|<\infty .$ When this holds, $I=\underset{(a,b)\to (-\infty ,\infty )}{\lim }{\int }_a^{b}f$ exists and is finite.

Properties §

1. If $f,g$ are integrable and coincide a.e., then $\int f=\int g$.
2. Integrability holds under linear combination, and $\int \alpha f+\beta g=\alpha \int f+\beta \int g$.
3. Monotonicity holds, i.e. $f\le g⟹\int f\le \int g$
4. If $f\le g$, $\int f=\int g⟹f=g$ a.e.

Theorems §

1. (Beppo Levi monotonicity) Given a non-negative monotonically increasing sequence $\{f_n\}$ with supremum $f=\underset{n}{\sup }{f}_n$, we have that $\int f=\underset{n}{\sup }\int f_n$.
2. (Fatou’s lemma) If $f_n(x)\ge 0$ a.e., then $\int f\le \underset{n\to \infty }{\mathrm{liminf}}\int f_n={\lim }_{n\to \infty }\underset{k\ge n}{\inf }\int f_k$.
3. (Lebesgue’s dominated convergence) Suppose $|f_n(x)|\le g(x)$ a.e. for some integrable $g$. Then, $\int f_n\underset{n\to \infty }{\to }\int f$. Some examples:
   1. Let $f_n(x)=\{\begin{array}{l}1,\text{ if }x>n\\ 0,\text{ if }x\le n\end{array}$. Then, $f_n(x)\underset{n\to \infty }{\to }0$ for all $x\in \mathbb{R}$, so $\int f=0$, but $\int f_n=\int 1_{[n,\infty )}=|[n,\infty )|=\infty$.
   2. Let $f_n(x)=\{\begin{array}{l}n,\text{ if }0<x<\frac{1}{n}\\ 0,\text{ otherwise }\end{array}$. Again, $f_n(x)\underset{n\to \infty }{\to }0$ for all $x\in \mathbb{R}$, so $\int f=0$, but $\int f_n=n{\int }_0^{\frac{1}{n}}dx=1$.

Remarks §

1. A bounded, measurable function is integrable.
2. The value of the Lebesgue integral $I$ does not depend on the specific limiting sequence $\{f_n\}$.
3. Any non-negative measurable function $f:\mathbb{R}\to [0,\infty )$ has a well-defined Lebesgue integral (possible infinite), and we say that $f$ is integrable exactly when the Lebesgue integral is finite.

Examples §

1. Consider $f(x)=\{\begin{array}{l}\frac{1}{x^{\alpha }},\text{ if }0<x<1\\ 0,\text{ otherwise}\end{array}$. Then, ${\int }_{-\infty }^{\infty }f(x)dx={\int }_0^{1}f(x)dx={\int }_0^1\frac{1}{x^{\alpha }}dx=\{\begin{array}{l}\frac{1}{1-\alpha },\text{ if }\alpha <1\\ \\ \infty ,\text{ if }\alpha \ge 1\end{array}.$Note that when $\alpha >0$, $f$ is unbounded and the Riemann integral cannot be used. Normally, we use an improper integral, but the Lebesgue integral provides an easier alternative.
2. Consider $f(x)=\{\begin{array}{l}1,\text{ if }x\in \mathbb{Q}\\ 0,\text{ otherwise}\end{array}$. Here, the Riemann integral does not exist, even over the interval $(0,1)$, but the Lebesgue integral is well-defined. In particular $|\mathbb{Q}|=0⟹f(x)=0$ a.e., so $\int f=\int 0=0$.