From an \(L^1\) Error to a Measure Bound
Suppose an integrable function \(f\) is approximated by a continuous function \(g\). The error \(f-g\) may be large at some points, and its averages over small balls may also be large. Smallness of the integral \(\int |f-g|\,dm\) does not give a pointwise bound. It does, however, control the measure of the set where averages of the error cross a fixed threshold. This is the useful bridge between approximation in \(L^1\) and local estimates.
Throughout, \(m\) denotes Lebesgue measure on \(\mathbb{R}^n\), and \(f,g\) are real-valued measurable functions. We take \(f\in L^1(m)\) and \(g\in C_c(\mathbb{R}^n)\), the continuous functions with compact support. In particular, \(e=f-g\) is integrable and locally integrable. We choose a representative of \(f\) that is finite everywhere, changing it on a null set if necessary; the integrals and maximal averages are unaffected by that change.
We use the weak-type estimate established earlier for this maximal function: if \(h\in L^1(m)\) and \(\lambda>0\), then $$ m\{Mh>\lambda\}\le \frac{5^n}{\lambda}\|h\|_1. $$ Here \(\|h\|_1=\int_{\mathbb{R}^n}|h|\,dm\). The estimate concerns the strict level set \(Mh>\lambda\); the constant \(5^n\) is a valid covering bound and need not be sharp. Measurability of \(Mh\) and the covering argument behind this estimate are part of the earlier material on maximal functions and Vitali covering.
Maximal Control of the Error
Proof. Since \(f\in L^1(m)\) and \(g\in C_c(\mathbb{R}^n)\), we have \(e=f-g\in L^1(m)\). Apply the weak-type estimate with \(h=e\). Since \(\|e\|_1=\|f-g\|_1\), it gives exactly the claimed inequality. If \(g_j\in C_c(\mathbb{R}^n)\) and \(\|f-g_j\|_1\to0\), then for any fixed \(\lambda>0\), $$ 0\le m\{M(f-g_j)>\lambda\}\le \frac{5^n}{\lambda}\|f-g_j\|_1\longrightarrow 0. $$ The threshold \(\lambda\) is fixed as \(j\) varies; the conclusion is convergence to zero of these measures, not a pointwise or uniform bound on the errors. \(\square\)
The estimate can also control local deviations from \(f(x)\), not just averages of \(e\). Define, for \(x\in\mathbb{R}^n\), $$ A_f(x)=\limsup_{r\downarrow0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y). $$ This quantity measures the upper limiting average deviation from the value at the center. It is measurable: for fixed \(r>0\), the ball average is measurable in \(x\), and its dependence on \(r\) is continuous for each \(x\) because \(f\) is locally integrable and ball boundaries have Lebesgue measure zero. Consequently, the suprema over ranges of radii can be taken over positive rational radii.
Proof. Fix \(x\in\mathbb{R}^n\) and \(r>0\). At each \(y\), write \(f(y)-f(x)=e(y)+(g(y)-g(x))+(f(x)-g(x))\). The triangle inequality, integration over \(B(x,r)\), and division by the ball's positive finite measure give $$ \frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)-f(x)|\,dm(y) \le \frac{1}{m(B(x,r))}\int_{B(x,r)}|e(y)|\,dm(y) +\frac{1}{m(B(x,r))}\int_{B(x,r)}|g(y)-g(x)|\,dm(y)+|e(x)|. $$ The first average is at most \(Me(x)\). By continuity of \(g\) at \(x\), for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(|g(y)-g(x)|<\varepsilon\) whenever \(|y-x|<\delta\). Thus the second average is at most \(\varepsilon\) whenever \(0<r<\delta\). Taking the limsup as \(r\downarrow0\), then letting \(\varepsilon\downarrow0\), proves \(A_f(x)\le Me(x)+|e(x)|\).
If \(A_f(x)>t\), then at least one of \(Me(x)>t/2\) and \(|e(x)|>t/2\) holds: otherwise their sum would be at most \(t\). Therefore $$ \{A_f>t\}\subseteq\{Me>t/2\}\cup\{|e|>t/2\}. $$ The weak-type estimate bounds the measure of the first set by \(2\cdot5^n\|e\|_1/t\). For the second set, the pointwise inequality \(|e|\ge (t/2)\chi_{\{|e|>t/2\}}\) gives $$ m\{|e|>t/2\}\le \frac{2}{t}\int_{\mathbb{R}^n}|e|\,dm. $$ Subadditivity of measure now yields \(m\{A_f>t\}\le 2(5^n+1)\|e\|_1/t\), as claimed. \(\square\)
Worked Estimates
Worked Example: The Maximal Function of an Interval Indicator
On \(\mathbb{R}\) with Lebesgue measure, let \(h=\chi_{[0,1]}\). Compute the strict superlevel set \(\{Mh>1/4\}\), then compare its measure with the weak-type bound.
If \(0<x<1\), sufficiently small centered intervals about \(x\) lie inside \([0,1]\), so \(Mh(x)=1\). At \(x=0\) and \(x=1\), centered intervals of radius \(r<1\) have half their length in \([0,1]\), so the maximal average is at least \(1/2\). For \(x<0\), an interval centered at \(x\) has zero intersection with \([0,1]\) if \(r\le -x\). If \(-x<r<1-x\), its intersection has length \(r+x\), and its average is \((r+x)/(2r)\), which increases with \(r\) because its derivative is \(-x/(2r^2)>0\). If \(r\ge1-x\), the average is \(1/(2r)\), which decreases with \(r\). Hence \(Mh(x)=1/(2(1-x))\). By reflection about \(1/2\), for \(x>1\) we have \(Mh(x)=1/(2x)\).
For \(x<0\), the strict inequality \(1/(2(1-x))>1/4\) is equivalent to \(x>-1\). For \(x>1\), \(1/(2x)>1/4\) is equivalent to \(x<2\). The points in \([0,1]\) also satisfy the strict inequality, including the endpoints. Therefore \(\{Mh>1/4\}=(-1,2)\), whose measure is \(2-(-1)=3\). The weak-type estimate gives $$ m\{Mh>1/4\}\le \frac{5}{1/4}\|h\|_1=20\cdot 1=20, $$ since \(\|h\|_1=m([0,1])=1\). The exact measure \(3\) is below this valid, non-sharp upper bound.
Worked Example: A Continuous Approximation with Explicit Error
Let \(f=\chi_{[0,1]}\) on \(\mathbb{R}\), and let \(\varepsilon=1/100\). Define \(g\) to be \(0\) for \(x\le0\), \(100x\) for \(0<x<1/100\), \(1\) for \(1/100\le x\le1\), \(101-100x\) for \(1<x<101/100\), and \(0\) for \(x\ge101/100\). This is continuous and compactly supported. Compute \(\|f-g\|_1\) and bound \(m\{A_f>1/2\}\).
On \([0,1/100]\), the error is \(1-100x\), whose integral is the area of a triangle with base \(1/100\) and height \(1\), namely \((1/2)(1/100)(1)=1/200\). On \([1,101/100]\), the error is \(101-100x\), a triangle with the same base and height, so its integral is also \(1/200\). The error vanishes elsewhere except at endpoints, which do not affect the integral. Thus \(\|f-g\|_1=1/200+1/200=1/100\).
Apply the theorem with \(n=1\) and \(t=1/2\). Its bound is $$ m\{A_f>1/2\}\le \frac{2(5+1)}{1/2}\cdot\frac{1}{100} =24\cdot\frac{1}{100}=\frac{6}{25}. $$ The right-hand side is a measure bound, not a claim that the exceptional set has exactly that measure. It also illustrates why an approximation with a small \(L^1\) error is useful: it makes the upper bound small at a fixed threshold.
Worked Example: Maximal Errors Along an Approximation Sequence
For each integer \(k\ge1\), take the preceding piecewise-linear construction with \(\varepsilon_k=2^{-k}\), and call its continuous function \(g_k\). Continue to let \(f=\chi_{[0,1]}\). Find an upper bound for \(m\{M(f-g_k)>1/4\}\), and evaluate it at \(k=6\).
The error has two triangular pieces, each with base \(2^{-k}\) and height \(1\). Each triangle has area \((1/2)2^{-k}\), so \(\|f-g_k\|_1=2^{-k}\). The weak-type estimate in dimension one gives $$ m\{M(f-g_k)>1/4\}\le \frac{5}{1/4}\,2^{-k}=20\cdot2^{-k}. $$ At \(k=6\), \(2^{-6}=1/64\), so the bound is \(20/64=5/16\). For general \(k\), \(20\cdot2^{-k}\to0\) as \(k\to\infty\). Hence the measures of these strict superlevel sets tend to zero at the fixed threshold \(1/4\). The estimate does not assert that \(M(f-g_k)\) tends to zero at every point.
What the Estimate Does—and Does Not—Say
The theorem combines two distinct controls. The maximal estimate bounds the measure where local averages of the error are large; the elementary integral bound controls the measure where the error at the center is large. Continuity of the approximant makes its local variation vanish as the ball shrinks. Together these give the stated control of \(A_f\). This is a useful approximation step in the study of differentiation, but the bound itself is not a claim that every point has convergent ball averages.
Common pitfall: confusing small integral with small values everywhere. For example, \(e=\chi_{[0,1/10]}\) on \(\mathbb{R}\) has \(\|e\|_1=1/10\), but \(e(x)=1\) for every \(x\in[0,1/10]\); in particular, \(|e(x)|\) is not bounded above by \(\|e\|_1\). For any \(x\in(0,1/10)\), sufficiently small centered balls lie inside that interval, so \(Me(x)=1\). Thus the incorrect inference “\(\|e\|_1=1/10\), therefore \(Me\le1/10\) everywhere” fails. The correct conclusion at threshold \(1/2\) is a measure bound: $$ m\{Me>1/2\}\le \frac{5}{1/2}\cdot\frac{1}{10}=1. $$ Indeed, the set contains \((0,1/10)\), of measure \(1/10\), but the estimate does not claim that its measure equals \(1\).
Check Your Understanding
In each question, keep the threshold fixed when comparing a family of approximation errors.
- If \(h\in L^1(m)\) on \(\mathbb{R}^n\), write the weak-type bound for the strict set \(\{Mh>\lambda\}\) and state the required condition on \(\lambda\).
- Suppose \(\|f-g\|_1=0.02\) on \(\mathbb{R}^2\). Use the stated weak-type estimate to bound \(m\{M(f-g)>0.1\}\).
- Explain why continuity of \(g\) makes the average of \(|g(y)-g(x)|\) over \(B(x,r)\) tend to zero as \(r\downarrow0\).
- In the local oscillation estimate, why is the set \(\{A_f>t\}\) contained in the union of the two strict superlevel sets at height \(t/2\)?
- Give an example showing why a small \(L^1\) norm cannot be used as a uniform bound on a function.