When Does a Measurable Function Have a Finite Integral?
For a nonnegative measurable function, the Lebesgue integral is defined even when its value is \(+\infty\). A real-valued measurable function can take both positive and negative values, so defining its integral requires care: positive and negative contributions must not both be infinite. The appropriate condition is that the function’s absolute value have finite integral.
This tutorial defines integrable functions and their signed integrals, then proves that sums and scalar multiples of integrable functions remain integrable. The results use the additivity and monotonicity of the nonnegative integral established earlier in this course. Throughout, \((X,\mathcal{F},\mu)\) is a measure space, and the functions under discussion are finite-valued and measurable.
Positive and Negative Parts
For a real number \(a\), its positive part is \(\max(a,0)\), and its negative part is \(\max(-a,0)\). Applying these operations pointwise gives the positive and negative parts of a function.
The maximum and minimum operations preserve measurability, so the positive and negative parts are measurable. The displayed identities follow by considering the two possibilities \(f(x)\geq0\) and \(f(x)<0\). In either case, $$ \int_X |f|\,d\mu = \int_X f^+\,d\mu+\int_X f^-\,d\mu, $$ by additivity of the nonnegative integral. Thus, if \(f\) is integrable, both terms on the right are finite. The difference defining \(\int_X f\,d\mu\) is consequently a difference of finite real numbers and is well-defined.
Conversely, if both \(\int_X f^+\,d\mu\) and \(\int_X f^-\,d\mu\) are finite, then their sum is finite, so \(\int_X|f|\,d\mu<\infty\). This explains the role of the absolute value: it ensures that the positive and negative contributions are each finite, rather than permitting an invalid difference of two infinite quantities.
Worked Examples
Worked Example: A Signed Step Function
Use Lebesgue measure on \(X=[0,4]\), and define \(f=2\) on \([0,2]\) and \(f=-3\) on \((2,4]\). The intervals are measurable, so \(f\) is measurable. Its positive and negative parts are $$ f^+=2\mathbf{1}_{[0,2]},\qquad f^-=3\mathbf{1}_{(2,4]}. $$ Since each interval has length \(2\), the indicator integral formula gives $$ \int_X f^+\,d\mu=2\cdot2=4,\qquad \int_X f^-\,d\mu=3\cdot2=6. $$ Thus \(f\) is integrable, and $$ \int_X f\,d\mu=4-6=-2,\qquad \int_X|f|\,d\mu=4+6=10. $$ The signed integral records the net contribution; the absolute integral records the total size of both contributions.
Worked Example: A Decaying Function on the Real Line
On \(\mathbb{R}\) with Lebesgue measure, let \(f(x)=1/(1+x^2)\). This is a nonnegative measurable function, so \(|f|=f\). Using the antiderivative \(\arctan(x)\), for \(R>0\) we have $$ \int_{-R}^{R}\frac{1}{1+x^2}\,dx = \arctan(R)-\arctan(-R) = 2\arctan(R). $$ As \(R\to\infty\), this tends to \(\pi\). By the increasing-domain approximation for nonnegative integrals, the integral over \(\mathbb{R}\) is therefore \(\pi\). In particular, $$ \int_{\mathbb{R}}|f|\,d\mu=\pi<\infty, $$ so \(f\) is integrable and \(\int_{\mathbb{R}}f\,d\mu=\pi\). The domain has infinite measure, but the function still has a finite integral because it decays sufficiently rapidly.
Worked Example: Cancellation Does Not Make \(1/x\) Integrable
On \((-1,1)\) with Lebesgue measure, define \(f(x)=1/x\) for \(x\ne0\), and set \(f(0)=0\). This is a finite-valued measurable function. Its positive part equals \(1/x\) on \((0,1)\) and zero elsewhere. Its negative part equals \(1/|x|\) on \((-1,0)\) and zero elsewhere. Both have infinite integral. For example, for \(0<\delta<1\), $$ \int_\delta^1\frac{1}{x}\,dx=-\log(\delta), $$ which tends to \(+\infty\) as \(\delta\) decreases to zero. The same calculation after reflecting the interval gives \(\int_{-1}^0 1/|x|\,dx=\infty\). Hence $$ \int_{-1}^{1} f^+\,d\mu=\infty,\qquad \int_{-1}^{1} f^-\,d\mu=\infty, $$ and \(f\) is not integrable. Symmetric cancellation does not define its Lebesgue integral: the proposed difference would be \(\infty-\infty\), which is undefined.
Integrable Functions Form a Vector Space
Linearity of the integral is useful only when the functions involved actually have finite integrals. The next theorem verifies both integrability and linearity, without subtracting infinite quantities.
Proof. The pointwise triangle inequality gives $$ |\alpha f+\beta g| \leq |\alpha|\,|f|+|\beta|\,|g|. $$ The right-hand side is nonnegative and measurable. By monotonicity, additivity, and nonnegative homogeneity of the nonnegative integral, $$ \int_X|\alpha f+\beta g|\,d\mu \leq |\alpha|\int_X|f|\,d\mu+|\beta|\int_X|g|\,d\mu <\infty. $$ Therefore \(\alpha f+\beta g\) is integrable.
It remains to prove the integral identity. First consider addition. Set \(h=f+g\), which is integrable by the argument just given. The pointwise identities \(f=f^+-f^-\), \(g=g^+-g^-\), and \(h=h^+-h^-\) imply $$ h^+ + f^-+g^-=h^-+f^++g^+. $$ Indeed, subtracting the right-hand side from the left gives \(h^+-h^- - f^+ + f^- - g^+ + g^-=h-f-g=0\). Every term in this equality is nonnegative and has finite integral: \(f^\pm,g^\pm,h^\pm\) are bounded above by \(|f|,|g|,|h|\), respectively. Integrating the equality and rearranging finite real numbers yields $$ \int_X h^+\,d\mu-\int_X h^-\,d\mu = \left(\int_X f^+\,d\mu-\int_X f^-\,d\mu\right) + \left(\int_X g^+\,d\mu-\int_X g^-\,d\mu\right). $$ By the definition of the signed integral, this is \(\int_X(f+g)\,d\mu=\int_X f\,d\mu+\int_X g\,d\mu\).
For a scalar \(c\geq0\), the positive and negative parts of \(cf\) are \(c f^+\) and \(c f^-\), so nonnegative homogeneity gives \(\int_X cf\,d\mu=c\int_X f\,d\mu\). If \(c<0\), the positive and negative parts of \(cf\) are \(|c|f^-\) and \(|c|f^+\). Thus $$ \int_X cf\,d\mu = |c|\left(\int_X f^-\,d\mu-\int_X f^+\,d\mu\right) = c\int_X f\,d\mu. $$ Apply scalar homogeneity to \(\alpha f\) and \(\beta g\), and then apply the addition result. This proves the stated identity and completes the proof.
The Integral Is Controlled by the Absolute Integral
Proof. Write \(A=\int_X f^+\,d\mu\) and \(B=\int_X f^-\,d\mu\). Both are finite and nonnegative. The definitions give $$ \left|\int_X f\,d\mu\right| = |A-B| \leq A+B = \int_X|f|\,d\mu. $$ The inequality \(|A-B|\leq A+B\) holds for all nonnegative real numbers \(A\) and \(B\), so the proof applies regardless of which part is larger.
This bound says that cancellation can reduce the signed integral, but cannot make its magnitude exceed the total absolute contribution. It is also a useful check in calculations: a claimed signed integral larger in magnitude than \(\int_X|f|\,d\mu\) must be incorrect.
Why the Definition Matters
A finite domain does not by itself guarantee integrability. The function in the third example is defined on an interval of finite measure but has infinite positive and negative integrals near zero. Conversely, a function on a domain of infinite measure can be integrable, as the decaying function in the second example shows. The decisive question is whether \(\int_X|f|\,d\mu\) is finite, not whether the domain has finite measure or whether positive and negative values appear to cancel.
The positive and negative parts also explain why the signed integral is a finite real number for every integrable function. If both parts had infinite integral, the difference used to define \(\int_X f\,d\mu\) would have no meaning. Requiring integrability rules out that situation and makes the linearity theorem valid. Later results can therefore manipulate integrable functions as a class while keeping every integral and subtraction finite.
Check Your Understanding
Use the definitions and proofs above to answer these questions.
- Why does \(\int_X|f|\,d\mu<\infty\) imply that both \(\int_X f^+\,d\mu\) and \(\int_X f^-\,d\mu\) are finite?
- What prevents the function \(1/x\), with value zero assigned at the origin, from having a Lebesgue integral on \((-1,1)\)?
- Which pointwise inequality proves that a linear combination of integrable functions is integrable?
- Why are all terms finite when the positive-part identity in the proof of linearity is integrated?
- When can equality hold in \(\left|\int_X f\,d\mu\right|\leq\int_X|f|\,d\mu\), and what does equality suggest about cancellation?