Bounded Measurable Functions on Finite Measure Spaces

Why Boundedness Needs Finite Measure

A measurable function need not have a finite integral merely because its values are finite. The useful guarantee in this tutorial comes from combining two hypotheses: the function is bounded, and the whole space has finite measure. Together they give a constant integrable majorant for its absolute value. The resulting estimate is both a proof of integrability and a way to control the size of the integral.

Throughout, $(X,\mathcal{M},\mu)$ is a measure space with $\mu(X)<\infty$, unless another space is explicitly specified. Functions in the main theorem are real-valued and measurable everywhere on $X$. A real-valued function $f$ is bounded if there is a finite constant $M\ge 0$ such that $|f(x)|\le M$ for every $x\in X$. This is a pointwise condition, not just a bound outside a null set.

Definition: For a real-valued measurable function $f$, its positive and negative parts are $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Its signed Lebesgue integral is defined when at least one of $\int f^+\,d\mu$ and $\int f^-\,d\mu$ is finite; in that case it is $\int f^+\,d\mu-\int f^-\,d\mu$, with an infinite value allowed if the other part has infinite integral. The function is integrable if $\int |f|\,d\mu<\infty$; then its signed integral is finite.

Measurability ensures that $f^+$ and $f^-$ are measurable, but does not by itself make their integrals finite or even ensure that their difference is defined. The finite-measure theorem below gives the stronger conclusion of integrability.

The Bounded-Measurable Integrability Theorem

Theorem: Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$, and let $f:X\to\mathbb{R}$ be measurable. If $|f(x)|\le M$ for every $x\in X$, for some finite $M\ge0$, then $f$ is integrable and $$ \int_X |f|\,d\mu\le M\mu(X), \qquad \left|\int_X f\,d\mu\right|\le M\mu(X). $$

Proof. The function $|f|$ is nonnegative and measurable, and the pointwise bound gives $|f|\le M\chi_X$. By monotonicity of the nonnegative Lebesgue integral, and the integral formula for the constant simple function $M\chi_X$,

\int_X|f|\,d\mu\le\int_X M\chi_X\,d\mu=M\mu(X)<\infty.

Thus $f$ is integrable. In particular, $A=\int_X f^+\,d\mu$ and $B=\int_X f^-\,d\mu$ are finite, and the definitions give $\int_X f\,d\mu=A-B$ and $\int_X|f|\,d\mu=A+B$. Since $A,B\ge0$, $|A-B|\le A+B$. Therefore $\left|\int_X f\,d\mu\right|\le\int_X|f|\,d\mu\le M\mu(X)$, as required. If $M=0$ or $\mu(X)=0$, the same inequalities give zero, so these cases are included. $\square$

The first estimate controls the integral of the absolute value; the second controls the signed integral. The bound need not be attained. It depends on the chosen pointwise bound $M$, so a larger valid $M$ still gives a correct, possibly weaker, estimate.

Worked Example: A Signed Step Function on an Interval

Let $X=[0,2]$ with the Lebesgue sigma-algebra and Lebesgue measure $m$. Define $f(x)=3$ on $[0,\tfrac12)$ and $f(x)=-2$ on $[\tfrac12,2]$. These sets are measurable, disjoint, and cover $X$; thus $f$ is measurable. Its pointwise bound is $|f|\le3$, and $m(X)=2$.

By the integral formula for a simple function, the positive part has value $3$ on a set of measure $\tfrac12$, and the negative part has value $2$ on a set of measure $\tfrac32$. Hence

\int_X f^+\,dm=3\cdot\frac12=\frac32, \qquad \int_X f^-\,dm=2\cdot\frac32=3, \qquad \int_X f\,dm=\frac32-3=-\frac32.

Also, $\int_X|f|\,dm=\frac32+3=\frac92$, while the theorem gives $\int_X|f|\,dm\le3\cdot2=6$ and $\left|\int_Xf\,dm\right|\le6$. Check: the direct simple-function formula gives $3(\tfrac12)+(-2)(\tfrac32)=\tfrac32-3=-\tfrac32$, agreeing with the positive-negative-part calculation; its absolute value $\tfrac32$ is at most $6$.

Worked Example: Approximating the Integral of a Bounded Function

On $X=[0,1]$ with Lebesgue measure, let $f(x)=2x$. This function is measurable and satisfies $0\le f\le2$, so the theorem already implies integrability and $\int_X f\,dm\le2m(X)=2$. We compute its integral using an increasing sequence of simple functions and the Monotone Convergence Theorem.

For each integer $n\ge1$, set $N=2^n$. On $[k/N,(k+1)/N)$, for $k=0,\ldots,N-1$, define $g_n(x)=2k/N$, and set $g_n(1)=2$. Each $g_n$ is a nonnegative measurable simple function. Dyadic subdivision shows $g_n(x)\le g_{n+1}(x)\le2x$: each interval at level $n+1$ lies in one at level $n$, and its left endpoint is no smaller than the earlier left endpoint. On each such interval the difference $2x-g_n(x)$ is less than or equal to $2/N$, and $2/N\to0$. At $x=1$ the difference is zero. Therefore $g_n\uparrow f$ pointwise.

Each of the $N$ half-open intervals has measure $1/N$; the single endpoint $\{1\}$ has measure zero. Thus the simple-function integral is

\int_Xg_n\,dm =\sum_{k=0}^{N-1}\frac{2k}{N}\cdot\frac1N =\frac{2}{N^2}\cdot\frac{N(N-1)}2 =\frac{N-1}{N}.

The Monotone Convergence Theorem applies because the $g_n$ are measurable, nonnegative, and increase pointwise to $f$. Consequently, $\int_Xf\,dm=\lim_{n\to\infty}(N-1)/N=1$. Check: for $n=1$, $N=2$ and the formula gives $1/2$; for $n=2$, $N=4$ and it gives $3/4$. Both are below the theorem's bound $2m(X)=2$. Independently, the elementary antiderivative calculation $\int_0^1 2x\,dx=[x^2]_0^1=1$ agrees.

Two More Exact Integrals

Worked Example: A Bounded Function Supported on a Null Set

Let $X=[0,1]$ with Lebesgue measure, and let $Q_0=\mathbb{Q}\cap[0,1]$. The set $Q_0$ is countable and hence Borel measurable. Define $f=\chi_{Q_0}$. This is measurable, real-valued, and bounded by $1$; the theorem ensures it is integrable.

A countable subset of $\mathbb{R}$ has Lebesgue measure zero: each singleton has measure zero, and countable subadditivity bounds the measure of the union by the sum of those zero measures. Therefore $m(Q_0)=0$, and the simple-function formula gives

\int_X f\,dm=1\cdot m(Q_0)+0\cdot m(X\setminus Q_0)=0.

Check: $f^+=f$, $f^-=0$, so the positive-negative-part definition also gives $0-0=0$. The pointwise supremum of $f$ is $1$, but its integral is $0$; a pointwise bound alone does not determine the exact integral.

Worked Example: A Finite Weighted Measure Space

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be the power set of $X$, and define $\mu(E)$ as the sum of the weights of the points in $E$, with weights $\mu(\{a\})=\tfrac12$, $\mu(\{b\})=1$, and $\mu(\{c\})=\tfrac32$. This is a measure: disjoint sets contain disjoint collections of points, so their weights add. Its total mass is $\tfrac12+1+\tfrac32=3$. Define $f(a)=2$, $f(b)=-1$, and $f(c)=4$. Every function on this finite measurable space is measurable, and $|f|\le4$.

The positive part contributes $2\cdot\tfrac12=1$ at $a$ and $4\cdot\tfrac32=6$ at $c$; the negative part contributes $1\cdot1=1$ at $b$. Thus

\int_Xf\,d\mu=(1+6)-1=6, \qquad \int_X|f|\,d\mu=1+1+6=8.

The theorem's bounds are $\int_X|f|\,d\mu\le4\cdot3=12$ and $\left|\int_Xf\,d\mu\right|\le12$. Check: summing the signed value times the weight at each point gives $2(\tfrac12)+(-1)(1)+4(\tfrac32)=1-1+6=6$, the same exact integral.

What Can Fail Without the Hypotheses?

Finite measure matters. On $(\mathbb{R},\mathcal{L},m)$, the constant function $f(x)=1$ is measurable and bounded by $1$, but $\int_{\mathbb{R}}|f|\,dm=m(\mathbb{R})=\infty$. Boundedness alone does not imply integrability when the measure of the whole space is infinite.

Measurability alone does not define every signed integral. On $(\mathbb{R},\mathcal{L},m)$, set $f(x)=1$ for $x\ge0$ and $f(x)=-1$ for $x<0$. This real-valued function is measurable, but $\int f^+\,dm=m([0,\infty))=\infty$ and $\int f^-\,dm=m((-\infty,0))=\infty$. Writing $\int f\,dm=\infty-\infty=0$ is an invalid calculation: $\infty-\infty$ is undefined, so the signed integral is undefined. In contrast, on a finite measure space the theorem proves $\int|f|\,d\mu<\infty$, which makes both parts finite and the signed integral well-defined.

These conditions answer different questions. Measurability makes the positive and negative parts measurable; finiteness of at least one part's integral is needed for the signed integral to be defined; and integrability requires the stronger condition $\int|f|\,d\mu<\infty$. A finite measure space and a bounded measurable real-valued function guarantee that stronger condition.

Check Your Understanding

For each question, identify the measure space and verify that the stated hypotheses apply before calculating.

If $\mu(X)=5$ and $|f|\le3$ everywhere for a measurable real-valued $f$, what upper bound does the theorem give for $\int|f|\,d\mu$ and $\left|\int f\,d\mu\right|$?
On $[0,1]$ with Lebesgue measure, compute the integral of the function equal to $4$ on $[0,\tfrac14)$ and $-1$ on $[\tfrac14,1]$. Check it using positive and negative parts.
Why does boundedness of $f(x)=1$ not imply integrability on $\mathbb{R}$ with Lebesgue measure?
For a measurable real-valued function, what does $\int f^+=\int f^-=\infty$ imply about its signed integral?
In the finite weighted space with weights $\tfrac12,1,\tfrac32$, calculate the integral if the function has values $0,2,-2$, respectively.