Two Sources of Measure
Lebesgue measure assigns zero mass to every individual point. A measure that also includes weighted point masses behaves differently: the value of a function at an atom contributes to its integral, even if changing that value would have no effect on its Lebesgue integral. For a nonnegative simple function, the computation separates into a continuous contribution from Lebesgue measure and a finite sum of contributions at the atoms.
We work on the measurable space \((\mathbb{R},\mathcal{L})\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. Choose finitely many distinct points \(x_1,\ldots,x_N\in\mathbb{R}\) and weights \(c_1,\ldots,c_N>0\). The case \(N=0\) is allowed. For \(x\in\mathbb{R}\), the Dirac measure \(\delta_x\) is defined by \(\delta_x(E)=1\) if \(x\in E\), and \(\delta_x(E)=0\) if \(x\notin E\), for each \(E\in\mathcal{L}\). Define
This is a measure: for disjoint measurable sets, both \(m\) and each \(\delta_{x_j}\) are countably additive, and a finite sum of countably additive measures is countably additive. The measure \(\mu\) is not finite on \(\mathbb{R}\), since \(\mu(\mathbb{R})=\infty\), but it is sigma-finite: each interval \([-n,n]\) has measure at most \(2n+\sum_{j=1}^N c_j<\infty\), and these intervals cover \(\mathbb{R}\). It is also complete. Indeed, if \(\mu(E)=0\), then \(m(E)=0\), so every subset of \(E\) belongs to \(\mathcal{L}\); and no \(x_j\) belongs to \(E\), because \(c_j>0\). Thus every subset of \(E\) also has \(\mu\)-measure zero. In what follows, functions are real-valued, nonnegative, measurable simple functions, and integral equalities are exact, not merely almost everywhere.
The Mixed-Measure Computation Rule
Let \(\varphi\) be a nonnegative simple function and let \(E_a=\{x\in\mathbb{R}:\varphi(x)=a\}\) for each of its positive values \(a\). The sets \(E_a\) are measurable and pairwise disjoint. By the definition of the simple-function integral,
The first sum is the Lebesgue contribution. The second is the atomic contribution: each atom contributes its weight multiplied by the function’s value at that point. Since \(\varphi\) has only finitely many values and there are finitely many atoms, both sums are finite sums; a term \(a\,m(E_a)\) may nevertheless be infinite if \(m(E_a)=\infty\) and \(a>0\).
Proof. The positive level sets \(E_a=\{\varphi=a\}\) are disjoint measurable sets, and \(\varphi=0\) on the remaining level set. By the definition of the integral of a nonnegative simple function, \(\int\varphi\,d\mu=\sum_{a>0}a\mu(E_a)\). For each such \(E_a\), the definition of \(\mu\) gives \(\mu(E_a)=m(E_a)+\sum_{j=1}^N c_j\delta_{x_j}(E_a)\). Substituting this expression and distributing the finite sums yields $$ \int\varphi\,d\mu =\sum_{a>0}a\,m(E_a) +\sum_{j=1}^N c_j\sum_{a>0}a\,\delta_{x_j}(E_a). $$ For fixed \(j\), exactly one level set contains \(x_j\), including the zero level set if \(\varphi(x_j)=0\). Therefore \(\sum_{a>0}a\,\delta_{x_j}(E_a)=\varphi(x_j)\). Substitution proves the formula. If a Lebesgue term is infinite, the equality still holds because only finitely many nonnegative terms are being added. \(\square\)
Worked Example: Atoms Inside and at the Boundary of Intervals
Let \(\mu=m+3\delta_{1/2}+2\delta_1\) on \((\mathbb{R},\mathcal{L})\), and define \(\varphi=2\chi_{[0,1)}+\tfrac12\chi_{[1,2]}\). Compute \(\int\varphi\,d\mu\). The two intervals are disjoint: \(1\) is excluded from \([0,1)\) and included in \([1,2]\).
The positive level sets are \(E_2=[0,1)\) and \(E_{1/2}=[1,2]\), each of Lebesgue measure \(1\). Thus the Lebesgue contribution is $$ 2m([0,1))+\frac12m([1,2])=2\cdot1+\frac12\cdot1=\frac52. $$ At \(1/2\), the function value is \(2\), so the first atom contributes \(3\cdot2=6\). At \(1\), the function value is \(\tfrac12\), so the second contributes \(2\cdot\tfrac12=1\). Consequently, $$ \int\varphi\,d\mu=\frac52+6+1=\frac{19}{2}. $$ Check: Summing by level sets gives \(\mu([0,1))=1+3=4\) and \(\mu([1,2])=1+2=3\). Hence \(2\cdot4+\tfrac12\cdot3=8+\tfrac32=\tfrac{19}{2}\), in agreement.
Worked Example: A Point Can Have Zero Lebesgue Measure but Positive Mixed Measure
Let \(\mu=m+2\delta_0+5\delta_2\), and set \(\psi=3\chi_{\{0\}}+4\chi_{(0,1]}\). The sets \(\{0\}\) and \((0,1]\) are disjoint. The singleton has Lebesgue measure zero, while \(m((0,1])=1\). Compute the integral.
The Lebesgue part is $$ 3m(\{0\})+4m((0,1])=3\cdot0+4\cdot1=4. $$ The value at the atom \(0\) is \(\psi(0)=3\), giving the contribution \(2\cdot3=6\). The atom at \(2\) is outside the support, so \(\psi(2)=0\), giving \(5\cdot0=0\). Therefore \(\int\psi\,d\mu=4+6+0=10\).
Check: The level set where \(\psi=3\) is \(\{0\}\), of mixed measure \(\mu(\{0\})=0+2=2\). The level set where \(\psi=4\) is \((0,1]\), of mixed measure \(\mu((0,1])=1\). The level-set sum is \(3\cdot2+4\cdot1=10\). In particular, \(\{0\}\) is a Lebesgue null set but not a \(\mu\)-null set.
Worked Example: Several Step Levels and an Atom Outside the Support
Take \(\mu=m+2\delta_0+3\delta_1+5\delta_3\) and $$ \varphi=\frac12\chi_{[-1,0)}+2\chi_{[0,1]}+4\chi_{(1,2]}. $$ The three sets are disjoint, and each has Lebesgue measure \(1\). Compute the integral, taking care with the endpoint values.
The continuous contribution is $$ \frac12\cdot1+2\cdot1+4\cdot1=\frac12+2+4=\frac{13}{2}. $$ At \(0\), which belongs to \([0,1]\), the value is \(2\), so its contribution is \(2\cdot2=4\). At \(1\), also in \([0,1]\), the value is \(2\), so its contribution is \(3\cdot2=6\). The point \(3\) is outside all three sets, so its contribution is \(5\cdot0=0\). Therefore $$ \int\varphi\,d\mu=\frac{13}{2}+4+6+0=\frac{33}{2}. $$ Check: The mixed measures of the three level sets are \(\mu([-1,0))=1\), \(\mu([0,1])=1+2+3=6\), and \(\mu((1,2])=1\). Weighting those level-set measures gives \(\tfrac12\cdot1+2\cdot6+4\cdot1=\tfrac12+12+4=\tfrac{33}{2}\), as required.
Overlaps, Null Sets, and a Common Error
When a simple function is given as a sum of indicator functions, the value at a point in more than one set is the sum of all the corresponding coefficients. For Lebesgue measure, a discrepancy at a single endpoint does not change the integral, because a singleton has measure zero. For a mixed measure, that same endpoint may be an atom, so its exact function value matters.
For a concrete calculation, let \(\nu=m+4\delta_1\) and \(h=2\chi_{[0,1]}+3\chi_{[1,2]}\). At \(x=1\), both indicators equal \(1\), so \(h(1)=2+3=5\). Away from \(1\), the two intervals contribute Lebesgue integrals \(2\cdot1=2\) and \(3\cdot1=3\), respectively. The continuous contribution is \(2+3=5\), and the atomic contribution is \(4\cdot5=20\). Thus \(\int h\,d\nu=25\).
Common mistake: treating the two intervals as disjoint at \(1\) and assigning the atom only the first coefficient gives the incorrect value \(5+4\cdot2=13\). The correction is to evaluate the original function at the atom: \(h(1)=5\), not \(2\). Equivalently, split the overlapping representation into disjoint level sets: \(h=2\) on \([0,1)\), \(h=5\) on \(\{1\}\), and \(h=3\) on \((1,2]\). The Lebesgue contribution remains \(2\cdot1+3\cdot1=5\), while the atom lies in the level set of value \(5\), contributing \(4\cdot5=20\).
The practical method is therefore to identify the positive level sets, compute their Lebesgue measures, and separately evaluate the function at every listed atom. A point outside the support contributes zero; an atom on a boundary contributes according to the actual membership convention in the sets. In particular, one must not discard point values merely because they lie in Lebesgue null sets. The integral here is defined with respect to \(\mu\), not just \(m\).
Check Your Understanding
For each question, identify the level sets and the values at all atoms before computing.
- Let \(\mu=m+2\delta_0\) and \(f=3\chi_{[-1,0)}+\chi_{[0,2]}\). Compute \(\int f\,d\mu\), noting which interval contains \(0\).
- For \(\mu=m+4\delta_2\), compute the integral of \(g=5\chi_{[0,2]}+2\chi_{[2,3]}\). What is \(g(2)\)?
- Suppose \(\mu=m+c\delta_x\) with \(c>0\), and \(\varphi=a\chi_E\) for \(a>0\) and \(E\in\mathcal{L}\). Write the integral in terms of \(m(E)\), \(c\), and whether \(x\in E\).
- Give an example of a set that has zero Lebesgue measure but positive measure under a mixed measure of the stated form.
- Explain why the formula for the integral uses \(\varphi(x_j)\), rather than a value chosen from one of the indicator terms in a representation of \(\varphi\).