The Lebesgue Integral: Integration with Respect to a Measure
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1 Shortcomings of the Riemann Integral
The Lebesgue integral arose from an attempt to generalize the Riemann integral so that limits and integration could be interchanged much more freely. Recall that if we have a sequence of functions \((f_n)\), defined on a bounded interval \([a,b]\), converging uniformly to \(f = \lim_{n\to \infty } f_n\), then
\[ \begin{align} \lim _{n\to \infty } \int _a^b f_n(x) dx = \int _a^b \lim _{n\to \infty } f_n dx \end{align} \tag{1}\]
If, however, \((f_n)\) converges to \(f\) pointwise, then the limit and the integral cannot be so simply interchanged, even if the sequence \((f_n)\) is monotone. For example, if \(\mathbb {Q} \cap [0,1] = \left\{ r_n\right\}_{n=1}^\infty , f_n = I_{\left\{ r_i\right\}_{i=1}^n}\), then \((f_n)\) converges to \(f = I_{\mathbb {Q} \cap [0,1]}\) pointwise, but not uniformly. Here \(\lim_{n\to \infty } \int_0^1 f_n(x) dx = \lim_{n\to \infty } 0 = 0\), while \(\int_0^1 f dx\) does not exist. If we want the limit under the integral sign to be valid in such cases, we must require that \(f\) be integrable, and that its integral equal \(0\).
The problem with the Riemann integral can be viewed from a somewhat different angle. From school geometry we know that the volume of a right cylinder of height \(h\), i.e., of a figure \(A \times [0,h], A \subset \mathbb {R}\), is the product of the length of the base \(A\) and \(h\). The Riemann integral was introduced in an attempt to numerically express the area under the graph of a function. Accordingly, the most natural requirement for such a functional is that its value on an indicator function \(I_{A}\) should equal the length of \(A\). If the set \(A\) is an interval, a closed interval, or a finite disjoint union of intervals and closed intervals, then the Riemann integral satisfies this requirement. In general, however, as the example above shows, this is not the case. Moreover, this fails if only because, when introducing the Riemann integral, the question of which subsets of the line have a length, which subsets of the line can be measured, is not even discussed. As is clear from this course, this question is far from trivial. At this point, however, we have precisely the rigorous machinery to answer this question. We have rigorously introduced the notions of measurable sets and measure (length), in particular, Lebesgue measure on \(\mathbb {R}^{n}\).
2 Natural Requirements for a New Integral
So, at this point we have the machinery for introducing and measuring the length (or area, volume, measure) of subsets of an arbitrary set \(X\) β a triple \((X, \mathcal{F}, \mu )\), consisting of the set itself, a sigma-algebra of its subsets (subsets that can be measured), and a measure \(\mu\). Let \(f,g,f_1, f_2, \dots : X \to \overline{\mathbb {R}}\) be functions from \(X\) to the extended real line. Let us write down natural requirements for the new integral \(\int \cdot \quad\):
For any indicator function \(f = I_{A}, A \in \mathcal{F}\) of any measurable set \(A\), the integral must exist 1 and its value must equal the measure of this set, i.e., \(\int f = \mu \left(A\right)\);
The integral must be linear:
Homogeneity of the integral: if \(c \in \mathbb {R}\) and \(\int f\) exists, then \(\int cf\) must exist and \(\int cf = c \int f\) must hold 2.
Additivity of the integral: if \(\int f, \int g\) exist and are not infinities of opposite sign, then \(\int h\) must exist and \(\int h = \int f + \int g\) must hold, where
\[ h(x) := \begin{cases} f(x)+g(x), & f(x)+g(x) \text{ is defined} \\ 0, & \text{otherwise} \end{cases} \]
- The limit transition under pointwise monotone convergence, bounded below, must hold. That is, if \(0 \leq f_1 \leq f_2 \leq \dots\) and \(\int f_i\) exists \(\forall i\), then \(\int f\) must exist and \(\int f = \lim_n \int f_n\) must hold, where \(f := \lim_n f_n = \sup_n f_n\).
This statement is called Leviβs monotone convergence theorem. Note that later it will turn out that the integral of any nonnegative measurable function must always exist. Accordingly, the requirement that \(\int f_i \; \forall i\) exist will turn out to be superfluous.
The functional called the Lebesgue integral is introduced on the basis of these requirements.
Although the Lebesgue integral arose from the Riemann integral, in general it is constructed for measurable functions defined on an arbitrary measure space \((X, \mathcal{F}, \mu )\).
3 Construction of the Lebesgue Integral
3.1 Step 1. Nonnegative Simple Functions
Given requirements and , for any nonnegative simple function \(f = \sum_{i=1}^n c_i I_{A_i}, 0 \leq c_i\), the integral must exist and must equal
\[ \mathfrak {I}^{1}\left(f\right) := \sum _{i=1}^n c_i \cdot \mu \left(A_i\right) \]
(indeterminate forms of the type \(0 \cdot +\infty\) are taken to equal \(0\)).
For now we take only nonnegative simple functions, in order to avoid indeterminate forms of the type \((+\infty ) - (+\infty )\) in the definition.
The functional introduced clearly satisfies requirement . It is easy to prove that it also satisfies the linearity requirement (if the coefficient \(c\) is nonnegative). It satisfies this, in particular, because a [finite] linear combination of simple functions is always a simple function.
Considering requirement on monotone limit transitions leads to the following step.
3.2 Step 2. Nonnegative Measurable Functions
Let us now consider requirement for the integral. At this point the integral has been introduced for nonnegative simple functions. What kind of function \(f\) can be obtained as the pointwise limit of an increasing sequence of nonnegative simple functions \((f_n)\)? Clearly, such a function \(f\) must be measurable, since the pointwise increasing limit of measurable functions equals their pointwise supremum, and hence is measurable. It must also be nonnegative.
It turns out that these two requirements, nonnegativity and measurability, are sufficient for the function \(f\) to be obtainable as a pointwise increasing limit of simple functions. Indeed, let \(f\) be nonnegative and measurable. Then, setting
\[ \begin{align} f_n := & \sum _{i=0}^{n-1} \sum _{j=0}^{2^n-1}\left( i + \frac{j}{2^n} \right) \cdot I_{f \in \left[i + \frac{j}{2^n}, i + \frac{j+1}{2^n}\right)} + \\ & + n \cdot I_{f \in [n, +\infty ]} \end{align} \]
\(f_n\) converges to \(f\) monotonically from below at every point of the domain. If \(f: (X, \mathcal{F}, \mu ) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) is a measurable nonnegative function (it may take the value \(+\infty\)), then we must set
\[ \mathfrak {I}^{2}\left(f\right) := \lim _n \mathfrak {I}^{1}\left(f_n\right), \]
where the functions \((f_n)\) are defined above.
One can prove a stronger statement:
\[ \mathfrak {I}^{2}\left(f\right) = \sup \{ \mathfrak {I}^{1}\left(s\right): \; 0 \leq s \leq f, \; s \text{ -- simple}\} \]
If \(f\) is nonnegative simple, then \(\mathfrak {I}^{2}\left(f\right) = \mathfrak {I}^{1}\left(f\right)\), i.e., \(\mathfrak {I}^{2}\left(\cdot \right)\) is an extension of \(\mathfrak {I}^{1}\left(\cdot \right)\) to a wider class of functions.
3.3 Step 3. Arbitrary Measurable Functions
At this point we have introduced the integral for nonnegative measurable functions. Let us return to considering the linearity requirement . Note that if \(f\) is measurable, then \(f = f^+ - f^-\), where
\[ f^+(x) := \max (f(x), 0), \quad f^-(x) := \max (-f(x), 0). \]
\(f^+, f^-\) are nonnegative measurable functions, hence, if at least one of the numbers \(\mathfrak {I}^{2}_{E}\left(f^+\right), \mathfrak {I}^{2}_{E}\left(f^-\right)\) is less than \(+\infty\), we must set
\[ \mathfrak {I}^{3}\left(f\right) := \mathfrak {I}^{2}\left(f^+\right) - \mathfrak {I}^{2}\left(f^-\right). \]
If \(f\) is nonnegative measurable, then \(\mathfrak {I}^{3}\left(f\right) = \mathfrak {I}^{2}\left(f\right)\). Thus, \(\mathfrak {I}^{3}\left(\cdot \right)\) is an extension of the functional \(\mathfrak {I}^{2}\left(\cdot \right)\) to a wider class of functions, namely, to the class of measurable functions for which \(\min (\mathfrak {I}^{2}\left(f^+\right), \mathfrak {I}^{2}\left(f^-\right)) < +\infty\).
The definition of this functional can be extended further still, for example, to functions defined \(\mu\)-a.e. on \(X\); however, for our course, what has already been constructed is sufficient. From now on we will denote \(\mathfrak {I}^{3}\left(f\right)\) by \(\int f d\mu\) and call it the Lebesgue integral. Possible alternative notations:
\[ \int _X f(x) d\mu (x), \quad \int f(x) \; \mu (dx), \quad \int _X f(x) \; \mu (dx). \]
Let us repeat once more: the Lebesgue integral \(\int f d\mu\) of a measurable function \(f\) exists if \(\min \left( \int f^+ d\mu , \int f^- d\mu \right) < +\infty\). A function is called Lebesgue integrable if its Lebesgue integral exists and is finite.
Let \(f, f_1, f_2, \ldots\) be measurable functions defined on the same measure space. We say that \((f_n)\) converges to \(f\) in \(L^p\), \(p \in (0, +\infty )\), if
\[ \int _{X}\left|f_n - f\right|^p \; d\mu \xrightarrow [n \to \infty ]{} 0 \]
4 Properties of the Lebesgue Integral
Throughout this section, \((X,\mathcal{F}, \mu )\) is a measure space, \(f,g, f_1, f_2, \dots : (X, \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) are measurable functions.
4.1 Linearity, Limit Properties
With this we have fulfilled the promise made in the chapter on limit theorems: Markovβs inequality has been proved in full generality β for an arbitrary measure space, not only for discrete distributions. Along with it, Chebyshevβs inequality is automatically established as well.
In particular, if \(\mu = \mathbb {P}\) is a probability measure, and \(\xi\) is a nonnegative random variable, then
\[ \mathbb {P}\left(\xi \geq t\right) \leq \frac{\mathbb {E}\left[\xi \right]}{t} \]
Markovβs inequality allows us to bound the βtailβ of the distribution of \(\xi\).
Markovβs inequality can be viewed slightly differently as follows:
\[ F_{\xi }(t) = \mathbb {P}\left(\xi \leq t\right) \geq \mathbb {P}\left(\xi < t\right) = 1 - \mathbb {P}\left(\xi \geq t\right) \geq 1 - \frac{\mathbb {E}\left[\xi \right]}{t} \]
Altogether,
\[ 1 \geq F_{\xi }(t) \geq 1 - \frac{\mathbb {E}\left[\xi \right]}{t} \]
We have obtained a lower bound for the distribution function of \(\xi\).
4.2 The Lebesgue Integral vs. the Riemann Integral
Up to this point we have considered the Lebesgue integral for a measurable function defined on an arbitrary measurable space \((X, \mathcal{F})\) with a measure \(\mu\). In this section we will discuss the Lebesgue integral for functions on \(\mathbb {R}^{k}\).
Let \(I = [\overrightarrow {a}, \overrightarrow {b}] = [a_1, b_1] \times \dots \times [a_k, b_k] \subset \mathbb {R}^{k}\) be a multidimensional rectangle.
Note that the theorem speaks of a function that is bounded on a closed interval (on a rectangle, in general). That is, it concerns the proper Riemann integral.