[proofplan] We show that the Lebesgue-Stieltjes integral $\int_0^t h(s)\, da(s)$ equals the limit of left-endpoint Riemann sums over any sequence of partitions with mesh tending to zero. The strategy is to approximate $h$ by step functions $h_m$ that are constant on each partition interval, express each Riemann sum as a Lebesgue-Stieltjes integral of $h_m$ against the signed measure $\mu$ induced by $a$, and then pass to the limit using the dominated convergence theorem. The left-continuity of $h$ ensures pointwise convergence $h_m \to h$, while the boundedness of $h$ provides the integrable dominator. The same argument applies to the total variation integral with $|\mu|$ replacing $\mu$. [/proofplan]

[step:Identify the signed measure $\mu$ induced by $a$ and rewrite the integral] Since $a: [0,t] \to \mathbb{R}$ is cadlag and in $BV[0,t]$, the [Bijection Between Signed Measures and Cadlag BV Functions](/theorems/2071) associates to $a$ a unique signed measure $\mu$ on $[0,t]$ satisfying $\mu([0,s]) = a(s)$ for all $s \in [0,t]$. The Lebesgue-Stieltjes integral is then \begin{align*} \int_0^t h(s)\, da(s) = \int_{(0,t]} h(s)\, \mu(ds). \end{align*} [/step]

[step:Construct step function approximations $h_m$ constant on partition subintervals]Fix a sequence of partitions $\pi_m = \{0 = t_0^{(m)} < t_1^{(m)} < \cdots < t_{n_m}^{(m)} = t\}$ with mesh $|\pi_m| \to 0$. For each $m$, define the step function \begin{align*} h_m: (0,t] &\to \mathbb{R} \\ s &\mapsto h(t_{i-1}^{(m)}) \quad \text{for } s \in (t_{i-1}^{(m)}, t_i^{(m)}], \quad i = 1, \ldots, n_m. \end{align*} Each $h_m$ is measurable with respect to $\mathcal{B}((0,t])$ since it is a finite linear combination of indicator functions of Borel sets.[/step]

[guided]The idea is to build a simple function that matches the Riemann sum structure. In the $m$-th Riemann sum, the integrand $h$ is evaluated at the left endpoint $t_{i-1}^{(m)}$ of each subinterval, and the increment $a(t_i^{(m)}) - a(t_{i-1}^{(m)})$ corresponds to the measure $\mu((t_{i-1}^{(m)}, t_i^{(m)}])$. So the step function $h_m$ is defined to take the constant value $h(t_{i-1}^{(m)})$ on the half-open interval $(t_{i-1}^{(m)}, t_i^{(m)}]$, exactly matching the structure of a left-endpoint Riemann sum. Formally, for each $m$, define \begin{align*} h_m: (0,t] &\to \mathbb{R} \\ s &\mapsto h(t_{i-1}^{(m)}) \quad \text{for } s \in (t_{i-1}^{(m)}, t_i^{(m)}], \quad i = 1, \ldots, n_m. \end{align*} Each $h_m$ is Borel measurable on $(0,t]$ since it is a finite linear combination of indicator functions of half-open intervals, which are Borel sets.[/guided]

[step:Express each Riemann sum as $\int_{(0,t]} h_m\, d\mu$] By construction, for each $m$, \begin{align*} \int_{(0,t]} h_m(s)\, \mu(ds) &= \sum_{i=1}^{n_m} h(t_{i-1}^{(m)}) \cdot \mu\bigl((t_{i-1}^{(m)}, t_i^{(m)}]\bigr) \\ &= \sum_{i=1}^{n_m} h(t_{i-1}^{(m)}) \bigl(a(t_i^{(m)}) - a(t_{i-1}^{(m)})\bigr), \end{align*} where the second equality uses the correspondence $\mu((s,u]) = a(u) - a(s)$ from the bijection between signed measures and cadlag BV functions. The right-hand side is precisely the $m$-th Riemann sum. [/step]

[step:Verify pointwise convergence $h_m(s) \to h(s)$ for all $s \in (0,t]$ using left-continuity of $h$] Fix $s \in (0,t]$. For each $m$, there exists a unique index $i_m$ such that $s \in (t_{i_m - 1}^{(m)}, t_{i_m}^{(m)}]$, and $h_m(s) = h(t_{i_m - 1}^{(m)})$. Since $t_{i_m - 1}^{(m)} < s \leq t_{i_m}^{(m)}$ and both $s - t_{i_m - 1}^{(m)}$ and $t_{i_m}^{(m)} - s$ are at most $|\pi_m|$, we have $t_{i_m - 1}^{(m)} \to s$ from below as $m \to \infty$. Since $h$ is left-continuous on $(0,t]$, this gives \begin{align*} h_m(s) = h(t_{i_m - 1}^{(m)}) \to h(s) \quad \text{as } m \to \infty. \end{align*}[/step]

[guided]Why does $t_{i_m - 1}^{(m)} \to s$? The point $s$ lies in the interval $(t_{i_m - 1}^{(m)}, t_{i_m}^{(m)}]$, so $0 < s - t_{i_m - 1}^{(m)} \leq t_{i_m}^{(m)} - t_{i_m - 1}^{(m)} \leq |\pi_m|$. Since $|\pi_m| \to 0$, we get $t_{i_m - 1}^{(m)} \uparrow s$. Crucially, $t_{i_m - 1}^{(m)} < s$ for every $m$ (since $s$ lies in the open left end of the interval), so the approach is strictly from the left. Now the hypothesis that $h$ is left-continuous enters: left-continuity at $s$ means precisely that $h(u) \to h(s)$ as $u \uparrow s$. Since $t_{i_m - 1}^{(m)} \uparrow s$, we conclude $h_m(s) = h(t_{i_m - 1}^{(m)}) \to h(s)$. What would go wrong without left-continuity? If $h$ had a left-discontinuity at $s$, then $h(t_{i_m - 1}^{(m)})$ would converge to $h(s^-)$, not to $h(s)$, and the Riemann sums would converge to the wrong value. This is why left-endpoint Riemann sums require a left-continuous integrand.[/guided]

[step:Apply the dominated convergence theorem to pass to the limit]Since $h$ is bounded, there exists $M > 0$ such that $|h(s)| \leq M$ for all $s \in [0,t]$. By construction, $|h_m(s)| = |h(t_{i_m - 1}^{(m)})| \leq M$ for all $s \in (0,t]$ and all $m$. The constant function $M$ is integrable with respect to the total variation measure $|\mu|$: \begin{align*} \int_{(0,t]} M\, |\mu|(ds) = M \cdot |\mu|((0,t]) = M \cdot V_a(t) < \infty, \end{align*} where $V_a(t)$ denotes the total variation of $a$ on $[0,t]$, which is finite since $a \in BV[0,t]$. We have established: - $h_m \to h$ pointwise on $(0,t]$ (from the previous step), - $|h_m| \leq M$ $|\mu|$-a.e. on $(0,t]$, - $M \in L^1((0,t], |\mu|)$. The dominated convergence theorem for signed measures (applied to the positive and negative parts of $\mu$ separately) yields \begin{align*} \lim_{m \to \infty} \int_{(0,t]} h_m(s)\, \mu(ds) = \int_{(0,t]} h(s)\, \mu(ds) = \int_0^t h(s)\, da(s). \end{align*} Combining with the identity from the third step, this gives the first claimed equality: \begin{align*} \int_0^t h(s)\, da(s) = \lim_{m \to \infty} \sum_{i=1}^{n_m} h(t_{i-1}^{(m)})\bigl(a(t_i^{(m)}) - a(t_{i-1}^{(m)})\bigr). \end{align*}[/step]

[guided]We apply the dominated convergence theorem to the sequence of integrands $h_m$ with respect to the signed measure $\mu$. The dominated convergence theorem for signed measures requires: 1. **Pointwise convergence:** $h_m(s) \to h(s)$ for $|\mu|$-almost every $s \in (0,t]$. We proved pointwise convergence for every $s$, which is stronger. 2. **Integrable dominator:** There must exist $g \in L^1((0,t], |\mu|)$ with $|h_m(s)| \leq g(s)$ for all $m$ and $|\mu|$-a.e. $s$. The constant $M = \|h\|_\infty$ works because $|h_m(s)| \leq M$ by construction, and \begin{align*} \int_{(0,t]} M\, |\mu|(ds) = M \cdot |\mu|((0,t]) = M \cdot V_a(t) < \infty. \end{align*} The finiteness of $V_a(t)$ holds because $a \in BV[0,t]$. Both conditions are verified, so the DCT gives \begin{align*} \lim_{m \to \infty} \int_{(0,t]} h_m(s)\, \mu(ds) = \int_{(0,t]} h(s)\, \mu(ds). \end{align*} The left-hand side equals $\lim_{m \to \infty} \sum_{i=1}^{n_m} h(t_{i-1}^{(m)})(a(t_i^{(m)}) - a(t_{i-1}^{(m)}))$ and the right-hand side is $\int_0^t h(s)\, da(s)$, establishing the first identity.[/guided]

[step:Deduce the total variation identity by the same argument with $|\mu|$ in place of $\mu$]The total variation measure $|\mu|$ is a positive finite Borel measure on $[0,t]$ with total mass $|\mu|((0,t]) = V_a(t)$. The total variation function $|a|(r) = V_a(r)$ satisfies $|\mu|((s,u]) = |a|(u) - |a|(s)$ for $0 \leq s < u \leq t$. Define step functions $h_m$ with respect to $|\mu|$ exactly as before. The integral of $h_m$ against $|\mu|$ is \begin{align*} \int_{(0,t]} h_m(s)\, |\mu|(ds) = \sum_{i=1}^{n_m} h(t_{i-1}^{(m)}) \cdot |\mu|\bigl((t_{i-1}^{(m)}, t_i^{(m)}]\bigr) = \sum_{i=1}^{n_m} h(t_{i-1}^{(m)}) \bigl(|a|(t_i^{(m)}) - |a|(t_{i-1}^{(m)})\bigr), \end{align*} where the second equality uses the correspondence $|\mu|((s,u]) = |a|(u) - |a|(s)$. Note that $|a|(t_i^{(m)}) - |a|(t_{i-1}^{(m)})$ is the total variation of $a$ on the subinterval $(t_{i-1}^{(m)}, t_i^{(m)}]$, which in general is $\geq |a(t_i^{(m)}) - a(t_{i-1}^{(m)})|$ (with equality only when $a$ is monotone on that subinterval). Applying the dominated convergence theorem to $(h_m)$ with respect to the positive measure $|\mu|$ (the same dominator $M$ works since $\int_{(0,t]} M\, |\mu|(ds) = M \cdot V_a(t) < \infty$) yields \begin{align*} \lim_{m \to \infty} \sum_{i=1}^{n_m} h(t_{i-1}^{(m)}) \bigl(|a|(t_i^{(m)}) - |a|(t_{i-1}^{(m)})\bigr) = \int_{(0,t]} h(s)\, |\mu|(ds) = \int_0^t h(s)\, |da(s)|. \end{align*} This establishes the second Riemann sum identity: the Riemann sums formed with total variation increments $|a|(t_i) - |a|(t_{i-1})$ converge to the integral against the total variation measure.[/step]

[guided]The second identity follows by repeating the entire argument with the total variation measure $|\mu|$ replacing the signed measure $\mu$. The key points carry over directly: 1. The step functions $h_m$ are unchanged. 2. The Riemann sum $\sum_i h(t_{i-1}^{(m)}) (|a|(t_i^{(m)}) - |a|(t_{i-1}^{(m)}))$ equals $\int_{(0,t]} h_m\, d|\mu|$ because $|\mu|((s,u]) = |a|(u) - |a|(s)$ is the total variation of $a$ on $(s,u]$. 3. Pointwise convergence $h_m \to h$ is the same as before. 4. The dominator $M$ is integrable with respect to $|\mu|$ since $\int_{(0,t]} M\, d|\mu| = M \cdot V_a(t) < \infty$. 5. The DCT for the positive measure $|\mu|$ yields the limit. An important subtlety: the Riemann sums here use the total variation increments $|a|(t_i) - |a|(t_{i-1})$, not the absolute values of the function increments $|a(t_i) - a(t_{i-1})|$. These coincide only when $a$ is monotone on each subinterval. The total variation increment $|a|(u) - |a|(s)$ accounts for all oscillation of $a$ within $(s, u]$, whereas $|a(u) - a(s)|$ only measures the net change. For the identity to hold as stated, the Riemann sums must be formed with the correct weights $|a|(t_i) - |a|(t_{i-1})$. Note that for positive measures the DCT is the standard version (no need to decompose into positive and negative parts), making this application slightly simpler than the signed-measure case.[/guided]