[proofplan] Set $h = f - g \in L^1_\mathrm{loc}(\Omega)$ with $\int_\Omega h\,\varphi \, d\mathcal{L}^n = 0$ for all $\varphi \in \mathcal{D}(\Omega)$. We show $h = 0$ a.e. by mollification: for any compact $K \subset \Omega$ and small $\varepsilon > 0$, the mollification $h * \rho_\varepsilon$ vanishes on the interior of $K$ (since translates of the mollifier are test functions), and $h * \rho_\varepsilon \to h$ in $L^1_\mathrm{loc}$ by the [Lebesgue Differentiation Theorem](/theorems/74). [/proofplan] [step:Reduce to showing $h = f - g$ vanishes a.e.] Suppose $T_f = T_g$, so $\int_\Omega (f - g)\,\varphi \, d\mathcal{L}^n = 0$ for all $\varphi \in \mathcal{D}(\Omega)$. Set $h = f - g \in L^1_\mathrm{loc}(\Omega)$. We must show $h = 0$ $\mathcal{L}^n$-a.e. on $\Omega$. [/step] [step:Show the mollification $h * \rho_\varepsilon$ vanishes on the interior of each compact set] Fix a compact set $K \subset \Omega$. Let $\rho_\varepsilon$ be a [standard mollifier](/page/Standard%20Mollifier) with $\operatorname{supp}(\rho_\varepsilon) \subseteq B(0, \varepsilon)$. For $\varepsilon$ sufficiently small and $x$ in the $\varepsilon$-interior $K_\varepsilon := \{x \in K^\circ : \operatorname{dist}(x, \partial \Omega) > \varepsilon\}$, the function $y \mapsto \rho_\varepsilon(x - y)$ belongs to $\mathcal{D}(\Omega)$ (it is smooth with support in $B(x, \varepsilon) \subset \Omega$). By hypothesis: \begin{align*} (h * \rho_\varepsilon)(x) = \int_\Omega h(y)\,\rho_\varepsilon(x - y) \, d\mathcal{L}^n(y) = 0 \quad \text{for all } x \in K_\varepsilon. \end{align*} [guided] The mollifier $\rho_\varepsilon(x - \cdot)$ is a test function for each fixed $x$ sufficiently far from $\partial\Omega$. This is the key point: the hypothesis $\int h\,\varphi \, d\mathcal{L}^n = 0$ for all $\varphi \in \mathcal{D}(\Omega)$ applies in particular with $\varphi(y) = \rho_\varepsilon(x - y)$. The left side is exactly the convolution $(h * \rho_\varepsilon)(x)$. For $x \in K_\varepsilon$, the ball $B(x, \varepsilon)$ is contained in $\Omega$, so $y \mapsto \rho_\varepsilon(x - y)$ has compact support in $\Omega$ and is smooth. Therefore $\rho_\varepsilon(x - \cdot) \in \mathcal{D}(\Omega)$, and the hypothesis gives $(h * \rho_\varepsilon)(x) = 0$. [/guided] [/step] [step:Conclude $h = 0$ a.e. via convergence of mollifications] Since $h \in L^1_\mathrm{loc}(\Omega)$, the [Lebesgue Differentiation Theorem](/theorems/74) gives $h * \rho_\varepsilon \to h$ in $L^1_\mathrm{loc}(\Omega)$ as $\varepsilon \to 0$. But $h * \rho_\varepsilon = 0$ on $K_\varepsilon$, and as $\varepsilon \to 0$ the sets $K_\varepsilon$ exhaust $K^\circ$. Therefore $h = 0$ $\mathcal{L}^n$-a.e. on $K^\circ$. Since $K \subset \Omega$ was an arbitrary compact set and $\Omega = \bigcup_j K_j^\circ$ for a compact exhaustion, $h = 0$ $\mathcal{L}^n$-a.e. on $\Omega$. That is, $f = g$ in $L^1_\mathrm{loc}(\Omega)$. [/step]