$\renewcommand{\R}{\mathbb{R}}$ # Lecture 1: The Lebesgue Integral and L1 ###### tags: 224a 2020 The Riemann integral has some drawbacks from the point of view of functional analysis, mostly related to the fact that it is not easy to characterize which functions are Riemann integrable, and it is not easy to tell when *pointwise limits of sequences* of Riemann integrable functions are integrable. For example, let $q_1,q_2\ldots$ be an enumeration of the rational numbers, and let $f_n:\mathbb{R}\rightarrow\mathbb{R}$ be the indicator function of the set $\{q_1,\ldots,q_n\}$, $n\in \mathbb{N}$. Then $f_n\uparrow f$, where $f$ is the indicator function of the rationals, and each $f_n$ is Riemann integrable since it has finitely many discontinuities; however, the monotone limit $f$ is not Riemann integrable (it is a nice exercise to check that the upper Riemann sums always converge to $1$ whereas the lower ones converge to $0$). The Lebesgue integral overcomes these drawbacks to a large extent. It is based on a simple idea: while the Riemann integral approximates the "area under a curve" by partitioning the *domain* of the integrand, the Lebesgue integral instead partitions the *range*. To formally define it we will need the notion of *Lebesgue measure* $m(\cdot)$, which is a consistent way of measuring the lengths of subsets of $\mathbb{R}$; see section I.3 of Reed and Simon for an introduction. For now we will note that $m(S)=0$ for every countable $S$, $m((a,b))=|a-b|$ for every interval, and $$m(\bigcup_n S_n)=\sum_n m(S_n) \qquad (*)$$ whenever $S_n$ is a countable collection of disjoint sets. Note that not all sets are measurable (i.e., are assigned a Lebesgue measure), but it is not easy to construct a non-measurable set and we will not come across them in this course. We will say that a statement is true almost everywhere (a.e.) if it is false on a subset of measure zero. Given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ let $S_f(t):=\{x: f(x)>t\}$ denote level sets of $f$. **Defn.** A function $f$ is *measurable* if $S_f(t)$ is measurable for every $t\in \mathbb{R}$. **Defn.** If $f$ is a nonnegative measurable function on $\mathbb{R}$, its *Lebesgue integral* is defined as $$\int_{\mathbb{R}} f(x)dx := \int_0^\infty m(S_f(t))dt,$$ where the integral on the RHS is an improper Riemann integral. If $f$ is an arbitrary function on $\mathbb{R}$ with decomposition $f=f_+-f_-$ for $f_+,f_-\ge 0$ and $\int |f(x)|dx<\infty$ then its Lebesgue integral is defined as $$\int_{\mathbb{R}} f(x)dx = \int_{\mathbb{R}} f_+(x)dx - \int_{\mathbb{R}} f_-(x)dx.$$ **Remark.** The domain of integration can be restricted to any measurable subset of $\R$ by defining $$\int_S f(x)dx := \int_S f(x) \mathbf{1}_S(x) dx,$$ where $\mathbf{1}$ is an indicator function. One can also easily extend the definition to complex valued functions on $\R$ and to functions defined on $\R^n$; see Reed and Simon for details. The beauty of the above construction is: 1. The new definition does not lose anything, i.e., every Riemann integrable function is also Lebesgue integrable (we will not prove this) 2. Many new functions are now integrable, e.g. the indicator function of any measurable set, including the rationals in the above example. 3. Familiar properties such as linearity and monotonicity of the integral as well as the triangle inequality are preserved (see Reed-Simon for details). ### The space L1 Let $$L^1_*(\mathbb{R}):=\{ f: \int |f(x)|dx<\infty\},$$ and define for functions in this set the *L1* norm to be $$\|f\|_1 := \int |f(x)|dx.$$ Note that $\|\cdot\|_1$ is not actually a norm as defined, since there are nonzero functions (such as the indicator of a finite set) for which it is zero. The space $L^1(\R)$ is defined as the set of equivalence classes of functions in $L^1_*$ with respect to the relation $f\sim g$ if $\|f-g\|_1=0$ , which happens for instance of $f$ and $g$ agree a.e. On this space it actually satisfies all of the axioms of a norm, some of which rely on the familiar properties mentioned above. We will routinely refer to "functions in L1" without explicitly talking about equivalence classes. We may similarly define spaces such as $L^1 (a,b)$ $L^1(\mathbb{C})$, and $L^1(\R^n)$ for functions with domains other than $\R$. ### Convergence Theorems The real payoff of the Lebesgue integral is its much better behavior under taking limits, which is encapsulated in the following two theorems. #### Monotone Convergence Theorem **Thm.** If $f_n\ge 0$, $\|f_n\|_1\le C$ for all $n$, and $f_n\uparrow f$ then $f\in L^1$ then $$\|f-f_n\|\rightarrow 0.$$ *Proof.* Fix $t>0$, and observe that for any $n$ we have by monotonicity: $$S_{f_n}(t)\subset S_{f_{n+1}}(t)\subset\ldots \subset S_f(t),$$ and the last set is measurable since it is a countable union of measurable sets. Property (*) applied to the (disjoint) differences $S_{f_{n+1}(t)}\setminus S_{f_n}(t)$ implies that $$\lim_{n\rightarrow \infty} m(S_{f_n}(t)) = m(S_f(t)).$$ Moreover, since for every $n$: $$t\cdot m(S_{f_n}(t))\le \|f_n\|_1 \le C,$$ the sequence $m(S_{f_n}(t))$ bounded, so it must have a finite limit and $m(S_f(t))<\infty$. We now have $$\lim_{n\rightarrow \infty}\int f_n(x)dx = \lim_{n\rightarrow\infty} \int_0^\infty m(S_{f_n}(t))dt = \int_0^\infty \lim_{n\rightarrow\infty} m(S_{f_n}(t))dt=\int f(x)dx,$$ where the second equality is by an elementary property of the Riemann integral (which is indeed well-behaved for monotone functions; details are left to the homework). The left hand side is convergent since $\|f_n\|\le C$, so we conclude that $f\in L^1$. Finally, since $f_n\le f$, we have by linearity: $$\lim_{n\rightarrow \infty} \int |f(x)-f_n(x)|dx = \lim_{n\rightarrow \infty} f_n(x)dx - \int f(x)dx=0,$$ as desired. Essentially, what happened in the above proof is that the question of exchanging limits and integrals was reduced to the case of *monotone* functions, for which nothing pathological can happen and it is easy to answer. The Monotone Convergence Theorem can be used to prove the even more useful **Dominated Convergence Theorem**; see Theorem I.11 of Reed and Simon for a statement and do it on the homework. From the above, it is a short step to the following pleasing conclusion, showing that $L^1$ is a complete metric space. #### Riesz-Fisher Theorem **Thm.** If $f_n$ is a Cauchy sequence in $(L^1(\R), \|\cdot\|_1)$ then there is a function $f\in L^1(\R)$ such that $\|f-f_n\|_1\rightarrow 0$. **Remarks** Another benefit of the Lebesgue integral is that it can easily be defined on spaces other than $\R^n$; see sec I.4 of Reed and Simon for details. The one thing that is lost in passing from the Riemann to the Lebesgue integral is that integrals no longer have an implicit *orientation*; this can be restored in several ways e.g. in the theory of differential forms.