The Lebesgue Integral

From Measurable Functions to an Integral

The previous tutorial studied convergence in measure and its relationship with almost-everywhere convergence. We now begin to assign a numerical size to a measurable function. The basic idea is to measure the sets on which the function takes particular values, first for simple functions and then for general nonnegative measurable functions.

Throughout this tutorial, let $(X,\mathcal{F},\mu)$ be a measure space. A simple function has finite range, as defined earlier in the course. For a nonnegative simple function, its positive values occur on finitely many disjoint measurable level sets. This makes it possible to define its integral by multiplying each value by the measure of the set where it occurs, then adding the results. Measures, and therefore integrals, are allowed to take the value $+\infty$.

Definition: Let $s:X\to[0,\infty)$ be a measurable simple function, with distinct positive values $a_1,\ldots,a_m$. Its integral is $$ \int_X s\,d\mu=\sum_{j=1}^{m}a_j\,\mu\bigl(\{x\in X:s(x)=a_j\}\bigr). $$ If $s$ is identically zero, the sum is empty and its integral is $0$. A term with positive coefficient and infinite measure is $+\infty$.

The zero set does not contribute to the sum. The formula depends only on the function’s values and their level sets: every representation of the same simple function has the same sets $\{x:s(x)=a_j\}$. Thus the definition is unambiguous, including when one or more of the level sets have infinite measure.

For a general nonnegative measurable function, there may be infinitely many values, so the finite sum above does not apply directly. Instead, use every nonnegative measurable simple function that lies below it. Their integrals provide lower estimates for the integral of the given function.

Definition: Let $f:X\to[0,\infty)$ be measurable. Its Lebesgue integral is $$ \int_X f\,d\mu = \sup\left\{\int_X s\,d\mu: s\text{ is a nonnegative measurable simple function and }s\leq f\right\}. $$ The supremum is taken in $[0,\infty]$. In particular, the integral may be infinite.

The zero function is among the simple functions beneath $f$, so this collection is nonempty. The definition asks for the best lower bound obtainable from simple functions, not necessarily for a single simple function that achieves the supremum. This distinction matters: a general measurable function need not itself be simple.

Basic Examples and Monotonicity

Worked Example: The Integral of an Indicator Function

Let $A\in\mathcal{F}$. The indicator $\mathbf{1}_A$ is a nonnegative simple function: it takes the value $1$ on $A$ and $0$ elsewhere. Its only positive level set is $A$. Therefore the simple-function definition gives

\int_X \mathbf{1}_A\,d\mu=\mu(A).

For example, on $\mathbb{R}$ with Lebesgue measure $\lambda$, take $A=(2,5]$. Its measure is $5-2=3$, so $$ \int_{\mathbb{R}}\mathbf{1}_{(2,5]}\,d\lambda=3. $$ The endpoints do not change this value: a bounded interval with endpoints $2$ and $5$, with any choice of included endpoints, has Lebesgue measure $3$, by the earlier theorem on the Lebesgue measure of bounded intervals.

The next result shows that comparing functions pointwise also compares their integrals. Its proof begins with simple functions, where a common finite partition lets us compare the defining sums.

Theorem (Monotonicity of the Lebesgue Integral): If $f,g:X\to[0,\infty)$ are measurable and $f\leq g$ on $X$, then $$ \int_X f\,d\mu\leq\int_X g\,d\mu. $$

Proof. First suppose that $s$ and $t$ are nonnegative measurable simple functions with $s\leq t$. Their finite ranges divide $X$ into finitely many measurable sets on each of which both functions are constant. More explicitly, intersect each level set of $s$ with each level set of $t$, including their zero level sets. These intersections form a finite measurable partition of $X$. On each partition set $C$, write $s=a$ and $t=b$. The pointwise inequality gives $0\leq a\leq b$, and hence $$ a\,\mu(C)\leq b\,\mu(C). $$ This remains valid if $\mu(C)=\infty$: when $a=0$ the left side is zero, and when $a>0$, also $b>0$, so both sides are infinite. Adding over the finite partition gives $\int_X s\,d\mu\leq\int_X t\,d\mu$.

Now let $f\leq g$ be the measurable functions in the theorem. Every nonnegative measurable simple function $s\leq f$ also satisfies $s\leq g$. Thus the collection of simple functions used to define $\int_X f\,d\mu$ is contained in the collection used to define $\int_X g\,d\mu$. Taking suprema over these collections proves $$ \int_X f\,d\mu\leq\int_X g\,d\mu. $$ This argument includes the case where either integral is infinite. $\square$

Worked Example: Integrating a Nonnegative Simple Function

On $[0,3]$ with Lebesgue measure, define $$ s(x)=2\mathbf{1}_{[0,1)}(x)+5\mathbf{1}_{[2,5/2]}(x). $$ The two intervals are disjoint. The function equals $2$ on a set of measure $1$, equals $5$ on a set of measure $1/2$, and equals $0$ elsewhere. Applying the simple-function definition,

\int_{[0,3]}s\,d\lambda =2\lambda([0,1))+5\lambda([2,5/2]) =2(1)+5(1/2) =\frac{9}{2}.

The value $0$ outside the two intervals contributes nothing. The calculation uses the measure of each level set, rather than the length of the entire interval $[0,3]$.

Signed Functions and Their Integrals

A real-valued measurable function may take both positive and negative values. Separate those parts by defining $$ f^+(x)=\max\{f(x),0\}, \qquad f^-(x)=\max\{-f(x),0\}. $$ Both functions are nonnegative and measurable, and $f=f^+-f^-$ pointwise. Their nonnegative integrals provide the definition for a signed function. If both integrals are infinite, their difference would have the undefined form $+\infty-\infty$, so no integral is assigned in that case.

Definition: For a measurable real-valued function $f$, if $\int_X f^+\,d\mu$ and $\int_X f^-\,d\mu$ are both finite, define $$ \int_X f\,d\mu=\int_X f^+\,d\mu-\int_X f^-\,d\mu. $$ More generally, if exactly one of these two nonnegative integrals is infinite, the extended integral is $+\infty$ when $\int_X f^+\,d\mu=+\infty$, and $-\infty$ when $\int_X f^-\,d\mu=+\infty$. If both are infinite, the extended integral is undefined.

A finite integral of a signed function requires both its positive and negative parts to have finite integrals. In particular, one cannot define the integral by subtracting two infinite quantities, even if informal cancellation seems plausible. Care with this issue is essential on spaces of infinite measure.

A Function Has Zero Integral Exactly When It Vanishes Almost Everywhere

For a nonnegative measurable function, the integral records whether positive values occur on sets of positive measure. The following result makes that statement precise. Here, “almost everywhere” means outside a measurable set of measure zero.

Theorem (Zero Integral Criterion): Let $f:X\to[0,\infty)$ be measurable. Then $$ \int_X f\,d\mu=0 $$ if and only if $f=0$ almost everywhere.

Proof. Suppose first that $\int_X f\,d\mu=0$. For each positive integer $n$, the set $$ E_n=\{x\in X:f(x)\geq 1/n\} $$ is measurable. The simple function $(1/n)\mathbf{1}_{E_n}$ lies below $f$ pointwise. By the definition of the integral and the indicator formula, $$ 0\leq \frac{1}{n}\mu(E_n) =\int_X \frac{1}{n}\mathbf{1}_{E_n}\,d\mu \leq\int_X f\,d\mu =0. $$ It follows that $\mu(E_n)=0$ for every $n$. Every point where $f(x)>0$ belongs to at least one $E_n$: choose $n$ large enough that $1/n\leq f(x)$. Therefore $$ \{x:f(x)>0\}=\bigcup_{n=1}^{\infty}E_n. $$ By Countable Subadditivity, this union has measure zero. Hence $f=0$ almost everywhere.

Conversely, suppose $f=0$ almost everywhere. There is a measurable null set $N$ such that $f(x)=0$ for every $x\notin N$. Let $s$ be any nonnegative measurable simple function with $s\leq f$. Then $s(x)=0$ outside $N$, so each positive level set of $s$ is contained in $N$ and has measure zero. The defining finite sum for $\int_X s\,d\mu$ is consequently zero. Every simple function in the supremum defining $\int_X f\,d\mu$ has integral zero, so that supremum is zero. This proves the converse. $\square$

Worked Example: A Nonzero Function with Zero Integral

On $[0,1]$ with Lebesgue measure, let $f=\mathbf{1}_{\mathbb{Q}\cap[0,1]}$. This function is not identically zero: it equals $1$ at every rational point in the interval. But $\mathbb{Q}\cap[0,1]$ is countable and hence has Lebesgue measure zero, by the earlier theorem that every countable subset of the real line is Lebesgue null. The indicator formula gives $$ \int_{[0,1]}f\,d\lambda =\lambda(\mathbb{Q}\cap[0,1]) =0. $$ Thus zero integral does not mean that a nonnegative function is zero at every point; it means that it is zero almost everywhere.

Approximating the Integral of the Identity Function

The definition by a supremum can be used even when the function is not simple. Consider $f(x)=x$ on $[0,1]$, with Lebesgue measure. For each positive integer $n$, put $m=2^n$. On the intervals $[k/m,(k+1)/m)$, where $k=0,\ldots,m-1$, define simple functions by $$ s_n(x)=\frac{k}{m}, \qquad t_n(x)=\frac{k+1}{m}. $$ At $x=1$, set $s_n(1)=t_n(1)=1$. These are measurable simple functions, and on each interval the endpoints give $$ s_n(x)\leq x\leq t_n(x). $$ The same inequalities hold at $x=1$.

Each interval has measure $1/m$, and the singleton $\{1\}$ has measure zero. Therefore $$ \int_{[0,1]}s_n\,d\lambda =\frac{1}{m^2}\sum_{k=0}^{m-1}k =\frac{m-1}{2m}, \qquad \int_{[0,1]}t_n\,d\lambda =\frac{1}{m^2}\sum_{k=0}^{m-1}(k+1) =\frac{m+1}{2m}. $$ By monotonicity, $$ \frac{m-1}{2m} \leq\int_{[0,1]}x\,d\lambda \leq\frac{m+1}{2m}. $$ As $n\to\infty$, $m=2^n\to\infty$, and both bounds tend to $1/2$. Hence $$ \int_{[0,1]}x\,d\lambda=\frac12. $$ The calculation exhibits the central construction: increasingly accurate simple lower bounds determine the integral through their supremum, while upper bounds can help identify its value.

What the Definition Does—and Does Not—Say

The Lebesgue integral begins with measurable level sets and their measures. For a simple function, it is a finite weighted sum. For a general nonnegative measurable function, it is the supremum of the integrals of all simple functions lying below it. This definition explains both why indicators integrate to measures and why changing a function on a null set does not change its integral.

A useful caution is that the definition of a nonnegative integral does not require a finite answer. A function can have an infinite integral, and the supremum need not be achieved by one of the simple functions beneath it. For signed functions, positive and negative parts must be handled separately; subtracting two infinite values is not permitted. The next tutorial develops the integral of simple functions and its basic calculations in greater detail.

Key takeaway: The Lebesgue integral of a nonnegative measurable function is the supremum of the integrals of the nonnegative simple functions beneath it. It is monotone, and it is zero exactly when the function vanishes almost everywhere.

Check Your Understanding

Use the definitions and proved results to check your understanding.

Why does the integral of an indicator function equal the measure of its set?
How does the supremum definition prove monotonicity when $f\leq g$?
Why is it invalid to define a signed integral as $+\infty-(+\infty)$?
In the zero integral criterion, why do the sets $\{f\geq 1/n\}$ cover $\{f>0\}$?
Why can a function that is nonzero at some points still have integral zero?

What You'll Learn