Local Averages and Pointwise Values
For an integrable function, a local average records what the function does around a point, not merely the value assigned at that point. This distinction is visible even for elementary step functions: at a jump, the averages can converge to a value different from the function’s value there. For indicators, the corresponding quantity is the proportion of a small interval occupied by the set. These computations explain how an individual point can be exceptional while the Lebesgue differentiation theorem still guarantees the expected behavior almost everywhere.
Throughout, \(m\) is Lebesgue measure on \(\mathbb{R}\), and \(f:\mathbb{R}\to\mathbb{R}\) is measurable and locally integrable, meaning \(\int_K |f|\,dm<\infty\) for every compact interval \(K\). For \(r>0\), define the centered local average at \(x\) by
The integral is finite under local integrability. Endpoints do not affect its value because singleton sets have Lebesgue measure zero. We say \(x\) is a Lebesgue point of \(f\) if \(\lim_{r\downarrow0}\frac{1}{2r}\int_{x-r}^{x+r}|f(t)-f(x)|\,dm(t)=0\). This is stronger than merely asking whether \(A_r f(x)\) has a limit: at a Lebesgue point that limit must equal the pointwise value \(f(x)\).
Continuity Points and Jump Points
Proof. Fix \(\varepsilon>0\). Continuity at \(x\) gives a \(\delta>0\) such that \(|f(t)-f(x)|<\varepsilon\) whenever \(|t-x|<\delta\). For \(0<r<\delta\), the triangle inequality for integrals yields
The same bound directly shows that the mean absolute deviation in the definition of a Lebesgue point is at most \(\varepsilon\). Since \(\varepsilon\) was arbitrary, both limits are zero in the required sense. \(\square\)
At a jump, the two sides contribute in equal proportion to a centered average. The function’s assigned value at the jump affects neither integral.
Proof. Given \(\varepsilon>0\), choose \(\delta>0\) such that \(|f(t)-L_-|<\varepsilon\) for \(x-\delta<t<x\) and \(|f(t)-L_+|<\varepsilon\) for \(x<t<x+\delta\). If \(0<r<\delta\), the left and right halves of the interval each have length \(r\). Therefore
This proves the limit. The value at \(x\) has no effect because changing an integrand at one point does not change its Lebesgue integral. \(\square\)
Worked Example: A Step Function at Its Jump
Let \(f(t)=2\) for \(t<0\) and \(f(t)=6\) for \(t\geq0\). Compute its average at the jump and compare it with the value there.
For any \(r>0\), the interval \((-r,r)\) contains a length \(r\) portion on which \(f=2\) and a length \(r\) portion on which \(f=6\). Thus
So the averages equal \(4\) for every radius, while \(f(0)=6\). The one-sided limits are \(L_-=2\) and \(L_+=6\), and their midpoint is \((2+6)/2=4\), as the jump proposition predicts. The point \(0\) is not a Lebesgue point: for every \(r>0\), the mean absolute deviation from \(f(0)=6\) is \(\bigl(4r+0r\bigr)/(2r)=2\), not a quantity tending to zero. Every \(x\ne0\) has a neighborhood on which \(f\) is constant, so its averages equal \(f(x)\) for all sufficiently small radii.
Indicator Averages and Density
Let \(E\in\mathcal{L}\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra on \(\mathbb{R}\). Its indicator \(\chi_E\) is \(1\) on \(E\) and \(0\) outside \(E\). Since \(\chi_E\) is bounded, it is locally integrable. Its local average is exactly the fraction of the interval lying in \(E\):
A point \(x\) is a density-one point of \(E\) if this fraction tends to \(1\) as \(r\downarrow0\); it is a density-zero point if the fraction tends to \(0\). The Density Theorem, established earlier in the course, says that for a Lebesgue measurable set \(E\), almost every point of \(E\) is a density-one point and almost every point of \(E^c\) is a density-zero point. The computations below use the definition directly.
Proof. The formula for \(A_r\chi_E(x)\) makes the first two assertions exactly the definitions of density one and density zero. For a density-one point with \(\chi_E(x)=1\), the absolute difference \(|\chi_E(t)-\chi_E(x)|\) equals \(1\) precisely on \(E^c\) and \(0\) on \(E\). Its average is \(m(E^c\cap(x-r,x+r))/(2r)=1-A_r\chi_E(x)\), which tends to zero. For a density-zero point with \(\chi_E(x)=0\), the absolute difference equals \(\chi_E(t)\), whose average is \(A_r\chi_E(x)\to0\). These are the Lebesgue point conditions. \(\square\)
Worked Example: Interior, Boundary, and Exterior of an Interval
Take \(E=(0,1)\). At \(x=\tfrac12\), if \(0<r<\tfrac12\), the entire interval \((x-r,x+r)\) lies in \(E\), so \(m(E\cap(x-r,x+r))=2r\) and \(A_r\chi_E(\tfrac12)=2r/(2r)=1\). This matches \(\chi_E(\tfrac12)=1\).
At \(x=0\), if \(0<r<1\), the part of \((-r,r)\) in \(E\) is \((0,r)\), of measure \(r\). Hence \(A_r\chi_E(0)=r/(2r)=\tfrac12\), whereas \(\chi_E(0)=0\). The same computation at \(x=1\) gives \(A_r\chi_E(1)=\tfrac12\), whereas \(\chi_E(1)=0\). Both endpoints are density points neither of density zero nor density one, and neither is a Lebesgue point of this indicator.
At \(x=2\), if \(0<r<1\), the interval \((2-r,2+r)\) misses \(E\), so the intersection has measure \(0\), and \(A_r\chi_E(2)=0=\chi_E(2)\). This check distinguishes an exterior point from a boundary point without relying on a picture.
Worked Example: A Density Point with the Wrong Pointwise Value
Let \(E=(-1,0)\cup(0,1)\), so that \(0\notin E\), and thus \(\chi_E(0)=0\). For every \(0<r<1\), the interval \((-r,r)\) lies in \(E\) except for the single point \(0\). Removing a singleton does not change Lebesgue measure, so \(m(E\cap(-r,r))=2r\) and \(A_r\chi_E(0)=2r/(2r)=1\).
Thus \(0\) is a density-one point of \(E\), but the chosen pointwise value is \(\chi_E(0)=0\). The averages converge to \(1\), not to that value; the absolute mean deviation from \(\chi_E(0)\) is \(1\) for every such \(r\). This is a genuine failure at the individual point \(0\), caused by excluding a point from a set that occupies essentially the whole neighborhood. It does not contradict the Density Theorem: a singleton has measure zero.
Exceptional Points: What Can and Cannot Happen
The Lebesgue Differentiation Theorem states that if \(f:\mathbb{R}\to\mathbb{R}\) is measurable and locally integrable, then for \(m\)-almost every \(x\), the mean absolute deviation from \(f(x)\) tends to zero. Consequently, \(A_r f(x)\to f(x)\) for almost every \(x\): indeed, the absolute difference between the average and \(f(x)\) is bounded by that mean absolute deviation. “Almost every” permits exceptional points, but only in a set of measure zero. In particular, under these hypotheses, failure of convergence to the pointwise value cannot occur on a set of positive measure.
The examples also show why the hypotheses matter. Continuity is sufficient for the desired limit at a point, but is not necessary: indicators can have the correct average at density points even when they are discontinuous there. Conversely, a mismatch at a boundary or at a deliberately excluded density point is a pointwise phenomenon. It must not be mistaken for failure on a positive-measure set.
Worked Example: Positive-Measure Failure When Local Integrability Is Dropped
Here is a finite-valued measurable function for which the local averages fail everywhere. This example drops local integrability, so it does not contradict the differentiation theorem. Enumerate the rationals as distinct numbers \(q_1,q_2,\ldots\), and consider the nonnegative measurable extended-valued function
where a summand is assigned the value \(+\infty\) at its rational singularity. To see that \(g\) is finite almost everywhere, define \(B_n=\{x:|x-q_n|<2^{-n/2}\}\). Its measure is \(m(B_n)=2\cdot2^{-n/2}\), and \(\sum_{n=1}^{\infty}m(B_n)=2\sum_{n=1}^{\infty}2^{-n/2}<\infty\). For every \(N\), subadditivity gives \(m(\bigcup_{n\geq N}B_n)\leq\sum_{n\geq N}m(B_n)\), whose right-hand side tends to zero. Therefore the set of points belonging to infinitely many \(B_n\) has measure zero. Outside that null set and the countable set of rationals, all but finitely many summands satisfy \(2^{-n}/|x-q_n|\leq2^{-n/2}\); the remaining tail is bounded by the convergent geometric series \(\sum_n2^{-n/2}\). The finitely many other summands are finite there.
Let \(N\) be the null set where \(g\) is infinite, enlarged to include all rationals, and define \(f(x)=g(x)\) for \(x\notin N\) and \(f(x)=0\) for \(x\in N\). Then \(f\) is measurable and finite-valued everywhere. Nevertheless, every nonempty open interval \(I\) contains some rational \(q_k\). For a sufficiently small \(a>0\), \((q_k-a,q_k+a)\subset I\), and the \(k\)-th summand has integral \(2^{-k}\int_{q_k-a}^{q_k+a}|x-q_k|^{-1}\,dm(x)=+\infty\). Since changing values on the null set \(N\) does not change the integral, \(\int_I f\,dm=+\infty\). In particular \(f\) is not locally integrable. For every \(x\in\mathbb{R}\) and every \(r>0\), \(A_r f(x)=+\infty\), while \(f(x)\) is finite. Thus convergence to \(f(x)\) fails at every point, including on sets of positive measure. The example isolates local integrability as essential to the almost-everywhere conclusion.
A reliable pointwise calculation therefore has three parts: identify the local geometry of the set or the one-sided behavior of the function, compute the normalized integral, and compare the limit with the pointwise value only after that computation. At continuity points, continuity controls the entire shrinking interval. At jumps, the two sides contribute equally. For indicators, density determines the average, while the value at the center can still be altered without changing any average.
Check Your Understanding
In each question, state whether the claimed limit is an average limit, a Lebesgue point conclusion, or both.
- For \(f(t)=3\) when \(t<2\) and \(f(t)=9\) when \(t\geq2\), compute \(A_r f(2)\) for every \(r>0\).
- For \(E=(0,1)\), compute \(A_r\chi_E(0)\) when \(0<r<1\). Is \(0\) a Lebesgue point of \(\chi_E\)?
- If \(x\) is a density-zero point of \(E\) but \(x\in E\), what is the limit of \(A_r\chi_E(x)\), and does it equal \(\chi_E(x)\)?
- Explain why changing a locally integrable function at one point can change its Lebesgue point status there without changing any of its local averages.
- Why does the positive-measure failure example not contradict the Lebesgue Differentiation Theorem?