Proof of Integration by Parts for Indefinite Lebesgue Integrals on a Compact Interval

Throughout, integrals over measurable subsets of $[0,T]$ are Lebesgue integrals with respect to the restricted Lebesgue measure; we use the linearity and monotonicity of the Lebesgue integral, together with additivity over disjoint measurable sets, which follows from linearity applied to the decomposition $f\,\mathbf{1}_{A\cup B}=f\,\mathbf{1}_A+f\,\mathbf{1}_B$ for disjoint measurable $A$ and $B$ (with $\mathbf{1}_A$ the function equal to $1$ on $A$ and $0$ off $A$).

Part (i). For $0\le s\le t\le T$, additivity gives $u_t-u_s=\int_{(s,t]}f(r)\,dr$, so $|u_t-u_s|\le\int_{(s,t]}|f(r)|\,dr$. Given $\varepsilon>0$, the absolute continuity of the Lebesgue integral applied to the integrable function $|f|$ yields $\delta>0$ such that $\int_A|f|\,dr\le\varepsilon$ whenever $A$ is a measurable subset of $[0,T]$ of measure less than $\delta$; since the interval $(s,t]$ has measure $t-s$, it follows that $|u_t-u_s|\le\varepsilon$ whenever $|t-s|<\delta$, which is continuity of $t\mapsto u_t$ on $[0,T]$. Moreover $|u_t|\le|u_0|+\int_{[0,T]}|f(r)|\,dr$ for every $t$, and likewise for $v$, so $C=\max\big(|u_0|+\int_{[0,T]}|f|,\ |v_0|+\int_{[0,T]}|g|\big)$ bounds both. The same argument gives continuity of $t\mapsto v_t$.

Part (ii): measurability and integrability. The map $t\mapsto v_t$ is continuous, hence measurable on $[0,T]$ with the trace Borel $\sigma$-algebra by measurability of sequentially continuous functions of measurable maps applied to the identity map $t\mapsto t$ (which is measurable, preimages of relatively open sets being relatively open) and the sequentially continuous function $v$ (continuity on $[0,T]$ implies sequential continuity). The products $t\mapsto f(t)v_t$ and $t\mapsto u_tg(t)$ are measurable by the same composition lemma, applied to the componentwise-measurable map $t\mapsto(f(t),v_t)$ (respectively $t\mapsto(u_t,g(t))$) and the sequentially continuous multiplication map on $\mathbb{R}^2$; and $|f(t)v_t|\le C|f(t)|$, $|u_tg(t)|\le C|g(t)|$ show they are integrable by comparison (monotonicity).

Part (ii): the identity. Consider the function $F(s,r)=f(s)\,g(r)$ on the product space $[0,T]\times[0,T]$, equipped with the product $\sigma$-algebra of two copies of the trace Borel $\sigma$-algebra and the product of two copies of the restricted Lebesgue measure; both factors are finite measure spaces (the restricted Lebesgue measure of $[0,T]$ is $T$ by the toolkit), hence $\sigma$-finite, so the Tonelli and Fubini theorems apply on the product. The factors $(s,r)\mapsto f(s)$ and $(s,r)\mapsto g(r)$ are product-measurable (preimages of level sets are measurable rectangles of the form $E\times[0,T]$ or $[0,T]\times E$), so $F$ is product-measurable, and by the Tonelli theorem

so $F$ is integrable on the product and the Fubini theorem applies to $F$ restricted to measurable subsets. The map $(s,r)\mapsto s-r$ is product-measurable, being the composition of the componentwise-measurable coordinate pair with the sequentially continuous subtraction map (composition lemma); hence $D_1=\{(s,r):s-r>0\}$ and its complement $D_2=\{(s,r):r\ge s\}$ are product-measurable, and $[0,T]^2=D_1\cup D_2$ disjointly. By the Fubini theorem, integrating first in $r$,

where $\int_{[0,s)}g=\int_{[0,s]}g=v_s-v_0$ because the singleton $\{s\}$ has measure zero. Similarly, integrating first in $s$,

On the other hand, by the Tonelli-Fubini computation above with $F$ in place of $|F|$ (valid since $F$ is integrable),

Combining the three displays,

Finally, expanding $u_Tv_T=\big(u_0+(u_T-u_0)\big)\big(v_0+(v_T-v_0)\big)$ and using $u_T-u_0=\int_{[0,T]}f$ and $v_T-v_0=\int_{[0,T]}g$ together with linearity of the integral,

which is the claimed identity.