The following question comes from a research paper. I have simplified some statements and backgrounds.
For any smooth compactly supported scalar function $u(x,t)$ and $v(x,t)$ defined on $(x,t) \in \mathbb{R}^2,$ and given a first of operator $L=\partial_t+\partial_x$, we denote its adjoint operator is $L^*=-\partial_t-\partial_x$, the classical integration by parts told us that $$ \int_{t_1}^{t_2} \int_{\mathbb{R}} Lu \cdot v \ dx dt=\int_{t_1}^{t_2} \int_{\mathbb{R}} u \cdot L^*v \ dx dt+\int_{\mathbb{R}} (uv)|_{t=t_2}dx-\int_{\mathbb{R}} (uv)|_{t=t_1}dx. $$ Now I am stating some properties we have already deduced. If $u \in L^2(\mathbb{R}^2), Lu \in L^2(\mathbb{R}^2)$ ($Lu$ is understood as a distribution), then for any $t_0 \in \mathbb{R},$ we have $u|_{t=t_0} \in H^{-1/2}(\mathbb{R})$ and the following estimate holds $$ \| u|_{t=t_0}\|_{H^{-1/2}(\mathbb{R})} \leq C(\| u\|_{L^2(\mathbb{R}^2)}+\| Lu \|_{L^2(\mathbb{R}^2)}). $$ Moreover, $\mathcal{D}(\mathbb{R}^2)$ (compactly supported functions) are dense in $\{(u,Lu): u \in L^2(\mathbb{R}^2),Lu \in L^2(\mathbb{R}^2) \}$ with respect to the graph norm $\| u\|_{L^2(\mathbb{R}^2)}+\| Lu\|_{L^2(\mathbb{R}^2)}.$
Hence, for $u,Lu \in L^2$, we know that there is a sequence of functions $u_n \in \mathcal{D}(\mathbb{R}^2)$ so that $u_n,L u_n$ converge to $u,Lu$ in $L^2(\mathbb{R}^2),$ and $u_n|_{t=t_0}$ converges to $u|_{t=t_0}$ in $H^{-1/2}(\mathbb{R})$ for any $t_0 \in \mathbb{R}.$
Clearly, for any compactly supported function $v$ and $u_n,$ the classical integration by parts formula holds: \begin{equation} \begin{aligned} \int_{t_1}^{t_2} \int_{\mathbb{R}} Lu_n \cdot v dt&=\int_{t_1}^{t_2} \int_{\mathbb{R}} u_n \cdot L^*v dt+\int_{\mathbb{R}} (u_nv)|_{t=t_2}dx-\int_{\mathbb{R}} (u_nv)|_{t=t_1}dx \\ &=\int_{t_1}^{t_2} \int_{\mathbb{R}} u_n \cdot L^*v dt+\langle u_n|_{t=t_2},v|_{t=t_2}\rangle_{\mathcal{D}',\mathcal{D}}-\langle u_n|_{t=t_1},v|_{t=t_1}\rangle_{\mathcal{D}',\mathcal{D}}. \end{aligned} \end{equation} The paper said that if we let $n\rightarrow +\infty$, then we have \begin{equation} \begin{aligned} \int_{t_1}^{t_2} \int_{\mathbb{R}} Lu \cdot v dt=-\int_{t_1}^{t_2} \int_{\mathbb{R}} u \cdot L^*v dt+\langle u|_{t=t_2-},v|_{t=t_2}\rangle_{\mathcal{D}',\mathcal{D}}-\langle u|_{t=t_1+},v|_{t=t_1}\rangle_{\mathcal{D}',\mathcal{D}}. \end{aligned} \end{equation} I am confused about the strict meaning of $u|_{t=t_1+}$ and $u|_{t=t_2-}$ as distribution here. It looks like the Lebesgue-Stieltjes integral, but I don't know how to deduce the expression. Can anyone help me? Any references are welcomed.
Lebesgue-Stieltjes integrals specify all Radon measure, or distributions of order 0. If $u|_{t=t_2-}$ were Radon measure, then setting $v=0$ near $t_1,$ one would need: \begin{equation} \begin{aligned} \int_{t_1}^{t_2} \int_{\mathbb{R}} Lu \cdot v dt -\int_{t_1}^{t_2} \int_{\mathbb{R}} u \cdot L^*v dt =\langle u|_{t=t_2-},v|_{t=t_2}\rangle_{\mathcal{D}',\mathcal{D}} \le C_K(u)\|v|_{t=t_2}\|_{K,\infty} \end{aligned} \end{equation} for some $C_K(u) > 0$ for every compact $K$ on $t=t_2,$ $\|v|_{t=t_2}\|_{K,\infty}=\sup_{x\in K}|v|_{t=t_2}(x)|.$ To check only smooth $u,$ one would require $C_K(u)$ to be bounded as functions of $\|u\|_2 + \|Lu\|_2.$
If one can provide estimation of $\int_K |u(x, t_2)|\:dx$ for smooth $u$ with compact support and compact $K$ uniformly using norm $\|u\|_2 + \|Lu\|_2,$ then assumption re. Radon measure is correct. It seems like $\int_K |u(x, t_2)|\:dx$ can be estimated as $(\int_K 1\:dx)\|Lu\|$ integrating along $x+t=const.$