From an Integrand to a Function
Given an integrable function \(g\), integrating from a fixed point to a variable endpoint produces a new function. Its value records the accumulated signed integral up to that endpoint. Even for simple integrands, computing this function requires attention to where the interval of integration begins and ends, and to the values of the integrand on each part of that interval.
Throughout, let \(a<b\), let \(\mathcal{L}_{[a,b]}\) be the Lebesgue sigma-algebra on \([a,b]\), and use the finite measure space \(([a,b],\mathcal{L}_{[a,b]},m)\), where \(m\) is Lebesgue measure. The integrands in this tutorial are real-valued and measurable, with \(g\in L^1(m)\), meaning \(\int_{[a,b]}|g|\,dm<\infty\). The integral is the signed Lebesgue integral. For \(x\in[a,b]\), the set \([a,x]\) is measurable and has finite measure.
The integrability assumption ensures that each value is finite: \(\left|\int_{[a,x]}g\,dm\right|\le\int_{[a,x]}|g|\,dm\le\int_{[a,b]}|g|\,dm<\infty\). If \(g\) is specified only as an element of \(L^1(m)\), it is an equivalence class of functions equal \(m\)-a.e.; the definition still gives one well-defined function \(F\).
Proof. Fix \(x\in[a,b]\). The function \(g-h\) is integrable and equals zero \(m\)-a.e. The Lebesgue integral of an integrable function that is zero a.e. is zero on every measurable set. Therefore $$ \int_{[a,x]}g\,dm-\int_{[a,x]}h\,dm =\int_{[a,x]}(g-h)\,dm=0. $$ This holds for each \(x\), so the two functions agree everywhere on \([a,b]\). \(\square\)
How the Endpoint Changes the Integral
Proof. The intervals \([a,x]\) and \((x,y]\) are disjoint and their union is \([a,y]\). Additivity of the integral over disjoint measurable sets gives $$ F(y)=\int_{[a,x]}g\,dm+\int_{(x,y]}g\,dm. $$ The singleton \(\{x\}\) has Lebesgue measure zero, so \(\int_{(x,y]}g\,dm=\int_{[x,y]}g\,dm\). Subtracting \(F(x)=\int_{[a,x]}g\,dm\) proves the formula, including \(x=y\), when both sides are zero. If \(g\ge0\) a.e., monotonicity of the integral gives \(\int_{[x,y]}g\,dm\ge0\), proving the last assertion. \(\square\)
This increment identity is a useful check on explicit formulas: the difference between the function's values at two endpoints must equal the integral of the original integrand between them. It also explains why a nonnegative integrand produces a nondecreasing function. No pointwise differentiability claim is needed to define or compute \(F\).
Worked Example: An Indicator of an Interval
On \(([0,5],\mathcal{L}_{[0,5]},m)\), set \(g(t)=\chi_{(1,3)}(t)\). This is measurable and integrable, since it is bounded and \(\int_0^5|g|\,dm=m((1,3))=2\). Compute \(F(x)=\int_{[0,x]}g\,dm\).
If \(0\le x\le1\), the intersection \([0,x]\cap(1,3)\) is empty, so \(F(x)=0\). If \(1\le x\le3\), that intersection has measure \(x-1\), including at \(x=1\) where its measure is zero. If \(3\le x\le5\), the entire interval \((1,3)\) is included, and its measure is \(2\). Thus $$ F(x)= \begin{cases} 0,&0\le x\le1,\\ x-1,&1\le x\le3,\\ 2,&3\le x\le5. \end{cases} $$
The formulas agree where their ranges meet: at \(x=1\), both give \(0\), and at \(x=3\), both give \(2\). Check: at \(x=4\), \(F(4)=2\), which is also the measure of \([0,4]\cap(1,3)\). The value assigned to the indicator at either endpoint \(1\) or \(3\) would not change any of these integrals, because singletons have Lebesgue measure zero.
Step Integrands Accumulate Piece by Piece
A step function is constant on each of finitely many intervals. Its indefinite integral is therefore computed by adding the contributions from the intervals already passed and the contribution from the part of the current interval up to \(x\). Values assigned to the step function at the finitely many partition points do not affect the integral.
Worked Example: A Step Function with Three Pieces
On \(([0,4],\mathcal{L}_{[0,4]},m)\), define \(g(t)=2\) on \([0,1)\), \(g(t)=-1\) on \([1,3)\), and \(g(t)=3\) on \([3,4]\). This measurable function is integrable; its absolute integral is \(2\cdot1+1\cdot2+3\cdot1=7\). Compute \(F(x)=\int_{[0,x]}g\,dm\).
For \(0\le x\le1\), the integral is \(2\) times the length \(x\), so \(F(x)=2x\). For \(1\le x\le3\), the first interval contributes \(2\cdot1=2\), and the part of the second interval contributes \(-1\cdot(x-1)=-(x-1)\). Hence \(F(x)=2-(x-1)=3-x\). For \(3\le x\le4\), the first two intervals contribute \(2\cdot1+(-1)\cdot2=0\), and the third contributes \(3(x-3)\). Therefore $$ F(x)= \begin{cases} 2x,&0\le x\le1,\\ 3-x,&1\le x\le3,\\ 3(x-3),&3\le x\le4. \end{cases} $$
Check: at \(x=1\), the first formula gives \(2\) and the second gives \(3-1=2\). At \(x=3\), the second gives \(3-3=0\) and the third gives \(3(3-3)=0\). At \(x=4\), the formula gives \(3(4-3)=3\); summing the full pieces independently gives \(2-2+3=3\), as required.
Power Functions, Including an Integrable Singularity
For \(p>-1\), the power function \(t^p\) is integrable on \([0,L]\) for any \(L>0\), when its value at \(0\) is assigned arbitrarily if necessary. The restriction \(p>-1\) matters: it makes the power \(p+1\) positive, which ensures that the contribution near zero is finite.
To justify the formula at the possibly singular endpoint, first take \(x>0\) and let \(n\) be large enough that \(1/n\le x\). The nonnegative measurable functions \(t^p\chi_{[1/n,x]}(t)\) increase as \(n\) increases and converge to \(t^p\) for every \(t\in(0,x]\), and to zero at \(t=0\). The Monotone Convergence Theorem applies: the functions are nonnegative and measurable, increase pointwise, and their limit is \(t^p\) a.e. on \([0,x]\). Thus $$ \int_{[0,x]}t^p\,dm =\lim_{n\to\infty}\int_{[1/n,x]}t^p\,dm =\lim_{n\to\infty}\frac{x^{p+1}-(1/n)^{p+1}}{p+1} =\frac{x^{p+1}}{p+1}. $$ The last limit is valid because \(p+1>0\). At \(x=0\), the integral is zero and the displayed formula also gives zero.
Worked Example: The Integrable Power \(t^{-1/2}\)
On \(([0,4],\mathcal{L}_{[0,4]},m)\), set \(g(t)=t^{-1/2}\) for \(t>0\) and \(g(0)=0\). This is measurable, and the power formula with \(p=-\tfrac12>-1\) gives $$ F(x)=\int_{[0,x]}g\,dm=\frac{x^{1/2}}{1/2}=2\sqrt{x},\qquad 0\le x\le4. $$ In particular, \(\int_{[0,4]}g\,dm=2\sqrt4=4\), so \(g\in L^1(m)\).
Check: for \(x>0\), direct integration away from zero gives \(\int_{[1/n,x]}t^{-1/2}\,dm=2\sqrt{x}-2/\sqrt{n}\). As \(n\to\infty\), this tends to \(2\sqrt{x}\), matching the formula. At \(x=4\), the result is \(4\), and at \(x=0\), the defining integral is zero, also matching \(2\sqrt0=0\).
A Common Error: Forgetting to Clip the Interval
For an indicator integrand, the integral is the measure of an intersection, not automatically the length of the full set indicated. For example, on \([0,5]\) with \(g=\chi_{(1,3)}\), the incorrect calculation \(F(4)=4-1=3\) treats the interval as though the integrand remained equal to \(1\) beyond \(3\). The correct calculation is $$ F(4)=m([0,4]\cap(1,3))=m((1,3))=3-1=2. $$
For a step function, the corresponding safeguard is to split the integral at every partition point and include only the portion of the current piece up to \(x\). The interval-increment formula provides a final check. These computations construct \(F\) and determine its values; they do not by themselves assert that its derivative equals \(g\) at every point.
Check Your Understanding
In each question, use the definition on the stated finite Lebesgue measure space and verify the endpoint values.
- On \([0,6]\), compute \(F(x)\) for \(g=\chi_{(2,5)}\) and check \(F(6)\) by measuring the intersection.
- On \([0,3]\), let \(g=1\) on \([0,1)\) and \(g=-2\) on \([1,3]\). Find the piecewise formula for \(F\), and check it at \(x=1\) and \(x=3\).
- For \(g(t)=t^{1/3}\) on \([0,8]\), compute \(F(x)\) and verify \(F(8)\).
- Explain why changing an integrable representative at finitely many points leaves its indefinite integral unchanged at every endpoint.
- For \(g(t)=t^p\) on \([0,1]\), what condition on \(p\) makes the power formula yield a finite integral at zero, and where is that condition used?