> [!definition] > > Let $f: \real^n \to \complex$ be a [[Measurable Function|measurable function]], then $f$ is **locally measurable** with respect to the [[Lebesgue Measure|Lebesgue measure]] if $\int_K\abs{f(x)}dx < \infty$ for every [[Compactness|compact]] set $K \subset \real^n$. The space of all locally integrable functions is denoted as $L^1_{\text{loc}}$, and is given a [[Topological Vector Space|topology]] induced by the following family of [[Seminorm|seminorms]]: > $ > \norm{f}_{K} = \norm{f \cdot \chi_K}_1 > $ > where $K$ is compact, which makes it a [[Fréchet Space|Fréchet space]]. > [!theorem] > > Let $f \in \loci$, $x \in \real^n$, and $r>0$. Define > $ > A_rf(x) = \frac{1}{m(B(x, r))}\int_{B(x, r)} f(y)dy > $ > as the average value of $f$ on $B(x, r)$. Then $A_rf(x): (0, \infty) \times \real^n \to \complex$ is [[Continuity|continuous]]. > > *Proof*. Let $\seq{(r_n, x_n)} \subset (0, \infty) \times \real^n$ be a convergent sequence with $(r_n, x_n) \to (r, x)$, then we have $\chi_{B(x_n, r_n)} \to \chi_{B(x, r)}$ pointwise. As both sequences are convergent, they are bounded and there exists $B(r_0, x_0)$ such that $\chi_{B(r_0, x_0)}$ dominates $\seq{\chi_{B(x_n, r_n)}}$. Since $f \in \loci$, $\chi_{B(r_0, x_0)} \cdot \abs{f} \in L^1$ dominates $\seq{\chi_{B(x_n, r_n)} \cdot \abs{f}}$. By the [[Dominated Convergence Theorem]], we have > $ > \int_{B(x_n, r_n)}f(y)dy \to \int_{B(x, r)}f(y)dy > $ > So the right half of the expression is continuous. On the other hand, $m(B(x, r)) = r^nm(B(0, 1))$ is a continuous expression of $r$. Therefore $A_rf(x)$ is continuous. > [!theorem] > > Let $f \in \loci$, then $\lim_{r \to 0}A_rf(x) = f(x)$ [[Almost Everywhere|a.e.]] In other words, > $ > \lim_{r \to 0}\frac{1}{m(B(x, r))} \int_{B(x, r)}[f(y) - f(x)]dy = 0 \quad (a.e.) > $ > > *Proof*. First suppose that $f$ is [[Continuity|continuous]] and let $\delta > 0$, then there exists $r > 0$ such that $\abs{g(y) - g(x)} < \delta$ for all $y \in B(x, r)$. In which case, > $ > \abs{A_rg(x) - g(x)} \le \frac{m(B(r, x))}{m(B(r, x))}\delta = \delta > $ > so $A_{r}g(x) \to g(x)$ as $r \to 0$. > > Now suppose that $f \in L^1$ and let $\varepsilon > 0$, then there exists $g \in C(\real^n) \cap L^1$ such that $\norm{f - g}_1 < \varepsilon$. Hence > $ > \begin{align*} > &\limsup_{r \to 0}\abs{A_rf(x) - f(x)} \\ > &= \limsup_{r \to 0}\abs{A_r(f - g)(x) + (A_rg - g)(x) + (g - f)(x)} \\ > &\le H(f - g)(x) + 0 + \abs{f - g}(x) > \end{align*} > $ > If > $ > E_\alpha = \bracs{x: \limsup_{r \to 0}\abs{A_rf(x) - f(x)} > \alpha} \quad F_\alpha = \bracs{x: \abs{f - g}(x) > \alpha} > $ > then > $ > E_\alpha \subset F_{\alpha / 2} \cup \bracs{x: H(f - g)(x) > \alpha/2} > $ > By the [[Hardy-Littlewood Maximal Function|maximal theorem]], there exists $C > 0$ such that > $ > m(\bracs{x: H(f - g)(x) > \alpha/2}) \le \frac{2C}{\alpha}\norm{f - g}_1 < \frac{2C\varepsilon}{\alpha} > $ > On the other hand, > $ > m(F_{\alpha/2}) \le \frac{1}{\alpha/2} \norm{f - g}_1 < \frac{2\varepsilon}{\alpha} > $ > Combining the two yields > $ > m(E_\alpha) \le \frac{2\varepsilon}{\alpha} + \frac{2C\varepsilon}{\alpha} > $ > Since $\varepsilon$ is arbitrary, we get that $m(E_\alpha) = 0$. As $\alpha$ is also arbitrary, choosing a sequence that goes to zero yields $m(E_0) = 0$. > > Lastly, if $f \in \loci$ is arbitrary, then choosing $f_n = f \cdot \chi_{B(0, n)} \in L^1$ yields that $A_rf(x) \to f(x)$ as $r \to 0$ for all $x$ with $\norm{x} < n$. As $n$ is arbitrary as well, the proposition holds almost everywhere. > [!theorem] > > Let $f \in \loci$. If $\int_U f d\mu = 0$ for all $U$ bounded open, then $f = 0$. > > *Proof*. Let $\nu$ be the signed measure defined as $E \mapsto \int_E f$, then by the Lebesgue differentiation theorem, > $ > f = \lim_{r \to 0}\frac{\int_{B(0, r)}f}{m(B(0, r))} = 0 \quad a.e. > $ > [!theorem] > > Let $f \in \loci$. If $\int f \phi = 0$ for all $\phi \in C_c^\infty$, then $f = 0$. > > *Proof*. Let $U$ be a bounded open set, and let > $ > U_\varepsilon = U \setminus \bracs{x: d(x, U^c) < \varepsilon} > $ > then $U_\varepsilon$ is [[Compactness|compact]]. Let $\phi_\varepsilon$ be a [[Smooth Bump Function|smooth bump function]] such that $\phi_{\varepsilon}|_{U_\varepsilon} = 1$ and $\phi_\varepsilon|_{U^c} = 0$. Since $f \in \loci$, $\int_U \abs{f} < \infty$ and by the dominated convergence theorem, $\int f\phi_\varepsilon \to \int f\chi_{U} = \int_U f$ as $\varepsilon \to 0$.