From Absolute Improper Convergence to a Lebesgue Integral
An improper Riemann integral is defined by taking limits of ordinary integrals over bounded intervals. A Lebesgue integral, by contrast, is defined on a measurable space and does not depend on endpoint truncations. Absolute convergence connects these two constructions: if the improper integral of \(|f|\) is finite, then \(f\) is Lebesgue integrable, and the two integrals of \(f\) agree.
Let \(I=(a,b)\) be an interval, where either endpoint may be infinite. Use the measure space \((I,\mathcal{L}(I),m)\), where \(\mathcal{L}(I)\) is the Lebesgue sigma-algebra restricted to \(I\), and \(m\) is Lebesgue measure. This space is complete and sigma-finite; it need not have finite measure. Suppose \(f:I\to\mathbb{R}\) is locally Riemann integrable, meaning that it is Riemann integrable on every compact interval contained in \(I\). In particular, \(f\) is measurable: each compact restriction is measurable by the standard equality theorem for bounded Riemann-integrable functions, and \(I\) is a countable union of compact intervals contained in \(I\).
Choose compact intervals \(K_n=[a_n,b_n]\subset I\) that increase to \(I\). The usual improper integral is obtained by taking limits as the truncation endpoints approach the endpoints of \(I\); if there are two improper endpoints, both endpoint limits are required. Absolute convergence means that the corresponding improper integral of the nonnegative function \(|f|\) has a finite value. Equivalently, along such an exhaustion, the ordinary integrals \(\int_{K_n}|f(x)|\,dx\) tend to a finite limit. Absolute convergence also ensures that changing the endpoint truncations does not change the limit: the difference between two truncations is bounded in absolute value by the integral of \(|f|\) over the portions of the endpoint tails between them.
Proof. Fix a compact exhaustion \(K_n=[a_n,b_n]\) of \(I\), and write \(A=\lim_{n\to\infty}\int_{K_n}|f(x)|\,dx<\infty\). For each \(n\), the function \(f\) is Riemann integrable on \(K_n\); so are \(|f|\), \(f^+=\max(f,0)\), and \(f^-=\max(-f,0)\). By the standard equality theorem for bounded Riemann-integrable functions, their Lebesgue integrals over \(K_n\) equal their Riemann integrals there.
The functions \(|f|\chi_{K_n}\) are measurable, nonnegative, and increase pointwise on \(I\) to \(|f|\), since \(K_n\) increase and their union is \(I\). The Monotone Convergence Theorem therefore applies and gives $$ \int_I |f|\,dm =\lim_{n\to\infty}\int_I |f|\chi_{K_n}\,dm =\lim_{n\to\infty}\int_{K_n}|f(x)|\,dx =A<\infty. $$ Thus \(f\) is Lebesgue integrable. More explicitly, \(0\le f^+\le |f|\) and \(0\le f^-\le |f|\), so both \(f^+\) and \(f^-\) have finite Lebesgue integrals.
Apply the Monotone Convergence Theorem again, separately to \(f^+\chi_{K_n}\) and \(f^-\chi_{K_n}\). It gives $$ \int_I f^+\,dm=\lim_{n\to\infty}\int_{K_n}f^+(x)\,dx, \qquad \int_I f^-\,dm=\lim_{n\to\infty}\int_{K_n}f^-(x)\,dx. $$ On each \(K_n\), linearity of the Riemann integral gives \(\int_{K_n}f=\int_{K_n}f^+-\int_{K_n}f^-\). Taking limits is valid because both sequences of positive integrals converge to finite values. Hence $$ \int_I f\,dm =\int_I f^+\,dm-\int_I f^-\,dm =\lim_{n\to\infty}\int_{K_n}f(x)\,dx. $$ The last limit is the improper Riemann integral by its definition, and absolute convergence makes it independent of the chosen endpoint truncations. This proves the asserted equality. \(\square\)
Examples at Endpoints and Infinity
Worked Example: A Singularity at the Left Endpoint
Let \(I=(0,1)\) and \(f(x)=x^{-1/2}\). Compute the absolute improper integral and identify the Lebesgue integral.
The function is nonnegative and continuous on every compact subinterval of \((0,1)\), so it is locally Riemann integrable. Since \(|f|=f\), truncate at \(\delta>0\):
As \(\delta\downarrow0\), this tends to \(2\), so the absolute improper integral converges. The theorem gives \(\int_{(0,1)}f\,dm=2\), equal to the improper Riemann value. Check: for the particular truncation \(\delta=1/16\), the integral is \(2-2(1/4)=3/2\). Independently, the substitution \(x=t^2\), \(dx=2t\,dt\), changes \(x^{-1/2}dx\) to \(2\,dt\) for \(0<t<1\), whose integral is \(2\).
Worked Example: A Signed Function on the Whole Line
On \(I=\mathbb{R}\), define \(f(x)=e^x\) for \(x<0\), and \(f(x)=-e^{-x}\) for \(x\ge0\). Show that the theorem applies and compute both integral values.
The function is continuous on either side of \(0\) and bounded on compact intervals; its single jump at \(0\) does not prevent local Riemann integrability. For \(R>0\), truncate to \([-R,R]\). The two signed contributions are $$ \int_{-R}^{0}e^x\,dx=1-e^{-R}, \qquad \int_0^R -e^{-x}\,dx=-(1-e^{-R}). $$ Thus \(\int_{-R}^R f(x)\,dx=0\). For the absolute value, the same two pieces are both positive: $$ \int_{-R}^R |f(x)|\,dx =(1-e^{-R})+(1-e^{-R}) =2(1-e^{-R}). $$ As \(R\to\infty\), this tends to \(2\). The theorem therefore gives \(\int_{\mathbb{R}}|f|\,dm=2\) and \(\int_{\mathbb{R}}f\,dm=0\), with the latter equal to the improper Riemann integral. Check: integrating over \([-R,0]\) and \([0,R]\) separately gives limits \(1\) and \(-1\); their sum is \(0\), while the corresponding absolute integrals have limits \(1\) and \(1\), summing to \(2\).
Worked Example: A Positive Tail on a Half-Line
Let \(I=(0,\infty)\) and \(f(x)=1/(1+x^2)\). Verify absolute convergence and compute the common integral.
This function is positive and continuous on \((0,\infty)\), so \(|f|=f\) and it is locally Riemann integrable. For \(R>0\), $$ \int_0^R |f(x)|\,dx =\int_0^R\frac{1}{1+x^2}\,dx =\arctan R-\arctan 0 =\arctan R. $$ The limit as \(R\to\infty\) is \(\pi/2\). Hence \(f\) is Lebesgue integrable and \(\int_{(0,\infty)}f\,dm=\pi/2\), equal to its improper Riemann integral. Check: at \(R=1\), the truncated value is \(\arctan 1=\pi/4\). For an independent evaluation of the limiting value, set \(x=\tan\theta\); then \(dx=(1+\tan^2\theta)d\theta\), so \(dx/(1+x^2)=d\theta\), and the limits \(x=0\) and \(x\to\infty\) correspond to \(\theta=0\) and \(\theta\to\pi/2\). The integral is again \(\pi/2\).
What Absolute Convergence Does—and Does Not—Say
The key estimate in the proof is not cancellation in \(\int f\), but finiteness of \(\int |f|\). That finiteness controls both the positive and negative parts, so their Lebesgue integrals are separately finite. It also controls the effect of changing truncations: the contribution from any omitted endpoint tail is bounded by the corresponding tail integral of \(|f|\).
A common error is to replace the integral of an absolute value by the absolute value of the integral. In the whole-line example above, that mistake would calculate \(\int_{\mathbb{R}}|f|\,dm\) as \(\left|\int_{\mathbb{R}}f\,dm\right|=|0|=0\). The direct computation gives \(2\), not \(0\): the positive and negative halves cancel in the signed integral, while both halves contribute positively to the absolute integral. In general, the valid inequality is \(\left|\int f\,dm\right|\le\int|f|\,dm\) when \(f\) is integrable; equality need not hold.
Also, convergence of the improper integral of \(f\) alone is not the hypothesis of this theorem. The theorem requires convergence of the improper integral of \(|f|\). The distinction matters because the Lebesgue integral of a real-valued function is finite only when its positive and negative parts both have finite integrals, equivalently when \(\int|f|\,dm<\infty\). Thus absolute improper convergence is precisely what licenses the conclusion here; it is not a claim that every convergent signed improper integral is Lebesgue integrable.
Check Your Understanding
For each question, keep the distinction between the signed integral and the integral of the absolute value explicit.
- For \(f(x)=x^{-1/3}\) on \((0,1)\), compute the truncated absolute integral and its limit. Which theorem then identifies the Lebesgue integral?
- In the whole-line example, compute the two separate limits for \(\int_{-R}^0 f(x)\,dx\) and \(\int_0^R f(x)\,dx\), and compare their sum with \(\int_{\mathbb{R}}|f|\,dm\).
- Why does the proof apply the Monotone Convergence Theorem to \(f^+\chi_{K_n}\) and \(f^-\chi_{K_n}\) separately, rather than subtracting two possibly infinite limits?
- State the precise hypothesis on the improper integral that is needed before concluding that a locally Riemann integrable real-valued function is Lebesgue integrable.
- For \(1/(1+x^2)\) on \((0,\infty)\), verify the truncated integral at \(R=1\) and state its limit as \(R\to\infty\).