From Interval Heights to Lebesgue Integrals
A step function on the real line is constant between finitely many breakpoints. Its integral is therefore calculated from the areas of finitely many constant pieces: multiply each height by the length of its interval, then add. To use this rule correctly, one must account for every interval exactly once, disregard the finitely many breakpoint values, and treat intervals of infinite length separately.
Throughout, \(m\) is Lebesgue measure on \(\mathbb{R}\), and the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra. This measure space is complete and sigma-finite, but not finite. All functions considered here are measurable and take values in \([0,\infty]\). For a nonnegative function, its Lebesgue integral is allowed to be \(+\infty\); it is integrable in the finite-integral sense when its integral is finite.
Each interval in this partition has a height \(c_j\geq0\) and a length \(\ell_j\), which is its usual endpoint difference when bounded and \(+\infty\) when unbounded. In the formula below, a zero height contributes zero even when the interval has infinite length: we define \(0\cdot\infty=0\) for this purpose. A positive height on an interval of infinite length contributes \(+\infty\).
The Interval-Area Formula
Proof. The intervals \(I_j\) are Lebesgue measurable, and the partition has only finitely many intervals. Away from the finite set of breakpoints, \(f\) agrees with the simple function \(\sum_j c_j\chi_{I_j}\). A finite set has Lebesgue measure zero, so changing a nonnegative measurable function on that set does not change its integral. By the definition of the integral of a nonnegative simple function, and additivity for its disjoint measurable pieces, \(\int_{\mathbb{R}}\sum_j c_j\chi_{I_j}\,dm=\sum_j c_jm(I_j)\). For a bounded interval \(m(I_j)\) is its endpoint difference. For an unbounded interval \(m(I_j)=+\infty\); the integral of the zero function there is zero, while a positive constant there has infinite integral. This gives the stated convention and formula. \(\square\)
The formula is independent of the particular partition used to describe the function. If a partition is refined by inserting extra breakpoints, each old interval is split into finitely many smaller intervals with the same height. Their lengths add to the old length when it is finite; if the old interval is unbounded, the pieces have total length \(+\infty\). Thus the contribution is unchanged, including the zero-height case. Any two finite partitions can be compared by refining both with all their breakpoints.
Proof. Each bounded partition interval has finite length, so its finite height contributes a finite amount. There are only finitely many such contributions, and their sum is finite. If the height is zero on each unbounded interval, those intervals contribute zero, proving that the total integral is finite. Conversely, if an unbounded interval has positive height \(c\), then \(c\,m(I)=+\infty\), so the whole nonnegative integral is infinite. Breakpoint values do not affect the integral by the theorem. \(\square\)
Worked Example: Three Constant Pieces
Let \(f\) have height \(3\) on \((-2,0)\), height \(1\) on \((0,4)\), and height \(0\) on the remaining open intervals. Assign any nonnegative values at the breakpoints. Compute \(\int_{\mathbb{R}}f\,dm\).
The first interval has length \(0-(-2)=2\), and the second has length \(4-0=4\). The other intervals have height zero. Applying the formula gives \[ \int_{\mathbb{R}}f\,dm=3\cdot2+1\cdot4+0=6+4=10. \] In particular, even if \(f(0)=+\infty\), that value is at a single point and does not alter the integral.
Check: The simple-function expression away from the breakpoints is \(3\chi_{(-2,0)}+\chi_{(0,4)}\). Its integral is \(3m((-2,0))+m((0,4))=3\cdot2+1\cdot4=6+4=10\), agreeing with the interval calculation.
Worked Example: Resolving Overlapping Intervals
Define \(f=2\chi_{(-3,2)}+5\chi_{(1,4)}\). Compute its integral. Both indicators are measurable, so \(f\) is a nonnegative measurable step function; on the overlap its height is the sum of the two heights.
The breakpoints are \(-3,1,2,4\). On \((-3,1)\), only the first interval contributes, so the height is \(2\), and the length is \(1-(-3)=4\). On \((1,2)\), both intervals contribute, so the height is \(2+5=7\), and the length is \(2-1=1\). On \((2,4)\), only the second contributes, so the height is \(5\), and the length is \(4-2=2\). The function is zero outside \((-3,4)\). Therefore \[ \int_{\mathbb{R}}f\,dm=2\cdot4+7\cdot1+5\cdot2=8+7+10=25. \]
Check: By additivity and positive homogeneity of the nonnegative integral, the original expression gives \(2m((-3,2))+5m((1,4))=2\cdot5+5\cdot3=10+15=25\). This agrees with the refined partition calculation. The overlap must be counted at height \(7\), not counted twice as two separate regions of height \(2\) and \(5\) without adding their contributions there.
Worked Example: A Positive Tail Gives an Infinite Integral
Let \(g\) be zero on \((-\infty,-2)\), equal to \(3\) on \((-2,1)\), and equal to \(1\) on \((1,\infty)\). Its values at \(-2\) and \(1\) may be arbitrary. Compute its integral.
The middle interval has length \(1-(-2)=3\), so it contributes \(3\cdot3=9\). The left interval contributes zero because its height is zero. The right interval has infinite length and positive height \(1\), so its contribution is \(+\infty\). Consequently, \[ \int_{\mathbb{R}}g\,dm=0+9+1\cdot(+\infty)=+\infty. \]
Check: For each positive integer \(n\), the interval \((1,1+n)\) lies in the region where \(g=1\), so \(\int_{\mathbb{R}}g\,dm\geq\int_{(1,1+n)}g\,dm=1\cdot n=n\). Since this holds for every positive integer \(n\), the integral cannot be finite. This confirms the infinite contribution from the positive tail.
Endpoints, Lengths, and Common Errors
A finite collection of endpoints has Lebesgue measure zero. Therefore, using open, closed, or half-open versions of the same partition intervals gives the same integral, as long as the function agrees on the intervals away from the endpoints. This does not mean endpoint values are absent from the definition of the function; it means their contribution to this Lebesgue integral is zero, even if a value at an endpoint is \(+\infty\).
A common mistake is to treat a breakpoint as if it contributed a positive-width piece. For the first worked example, one might incorrectly calculate \(3\cdot2+1\cdot4+f(0)\cdot1\), treating the single point \(0\) as an interval of length \(1\). If \(f(0)=7\), this wrong calculation gives \(10+7=17\). The correction is that \(\{0\}\) has Lebesgue measure zero: the integral is \(3\cdot2+1\cdot4+7\cdot m(\{0\})=6+4+7\cdot0=10\). A point value is not an additional interval height-length contribution.
A second error is to apply the finite-interval formula to an unbounded interval as if it had a finite length. The corollary gives the correct test: a positive constant on an unbounded interval makes the integral infinite; a zero constant there contributes zero. In particular, a step function may be finite-valued at every point and still have an infinite Lebesgue integral.
Check Your Understanding
Use the interval-area formula, and state how endpoint values or unbounded intervals affect each answer.
- A step function is \(4\) on \((-1,2)\), \(2\) on \((2,5)\), and zero elsewhere. What is its Lebesgue integral?
- For \(h=3\chi_{(0,2)}+2\chi_{(1,4)}\), partition the intervals at every breakpoint and compute \(\int_{\mathbb{R}}h\,dm\).
- Can changing a step function's value at one breakpoint from \(0\) to \(+\infty\) change its Lebesgue integral? Explain.
- A step function has height \(0\) on \((-\infty,0)\) and height \(2\) on \((0,3)\). Is its integral finite, and what is its value?
- A step function has height \(1/2\) on \((5,\infty)\) and is nonnegative everywhere else. What can be concluded about its integral?