[proofplan] We prove the identity by expanding the integral over the unit intervals on which the partial-sum function is constant. On each interval $[k,k+1)$ the value of $A(t)$ is the finite partial sum $A_k=\sum_{n=1}^k a_n$, so the integral against $f'$ becomes a finite sum of differences $A_k(f(k+1)-f(k))$. Rearranging this finite sum telescopes and leaves exactly the weighted sum $\sum_{n=1}^N a_n f(n)$. [/proofplan]

[step:Introduce finite partial sums and reduce the integral to unit intervals]For each integer $k$ with $1\le k\le N$, define \begin{align*} A_k:=\sum_{n=1}^k a_n. \end{align*} Then $A(N)=A_N$. If $N=1$, the integral over $[1,1]$ is $0$, and the claimed identity reduces to $a_1f(1)=A_1f(1)$. Assume now that $N\ge 2$. For each integer $k$ with $1\le k\le N-1$, the definition of $A$ gives $A(t)=A_k$ for every $t\in[k,k+1)$. Since changing a bounded function at finitely many points does not change its [Lebesgue integral](/page/Lebesgue%20Integral) with respect to $\mathcal L^1$, and since $A f'$ is integrable on $[1,N]$ because $A$ is bounded and $f'$ is continuous, we obtain \begin{align*} \int_1^N A(t)f'(t)\,d\mathcal L^1(t) = \sum_{k=1}^{N-1} A_k\int_k^{k+1} f'(t)\,d\mathcal L^1(t). \end{align*} For each such $k$, the one-dimensional [fundamental theorem of calculus](/theorems/632) for the continuously differentiable function $f$ gives \begin{align*} \int_k^{k+1} f'(t)\,d\mathcal L^1(t)=f(k+1)-f(k). \end{align*} Therefore \begin{align*} \int_1^N A(t)f'(t)\,d\mathcal L^1(t) = \sum_{k=1}^{N-1} A_k\bigl(f(k+1)-f(k)\bigr). \end{align*}[/step]

[guided]We introduce a notation that separates the jump sizes $a_n$ from the accumulated sums. For each integer $k$ with $1\le k\le N$, define \begin{align*} A_k:=\sum_{n=1}^k a_n. \end{align*} Thus $A_k$ is the value attained by the step function after the $k$-th jump, and in particular $A(N)=A_N$. First consider the endpoint case $N=1$. The interval $[1,1]$ has $\mathcal L^1$-measure $0$, so \begin{align*} \int_1^1 A(t)f'(t)\,d\mathcal L^1(t)=0. \end{align*} The asserted identity becomes \begin{align*} a_1f(1)=A_1f(1), \end{align*} which follows from the definition $A_1=a_1$. Now assume $N\ge 2$. The point of the proof is that $A$ is constant between consecutive integers. If $t\in[k,k+1)$ for an integer $k$ with $1\le k\le N-1$, then the integers $n$ satisfying $1\le n\le N$ and $n\le t$ are exactly $1,\dots,k$. Hence \begin{align*} A(t)=\sum_{n=1}^k a_n=A_k. \end{align*} The values of $A$ at the finitely many endpoints $2,\dots,N-1$ do not affect the Lebesgue integral, because finite subsets of $\mathbb R$ have $\mathcal L^1$-measure $0$. Also, $A$ is bounded because it takes only finitely many values, and $f'$ is continuous on the compact interval $[1,N]$, so the product $A f'$ is Lebesgue integrable. Therefore we may split the integral over the finite partition into unit intervals: \begin{align*} \int_1^N A(t)f'(t)\,d\mathcal L^1(t) = \sum_{k=1}^{N-1} A_k\int_k^{k+1} f'(t)\,d\mathcal L^1(t). \end{align*} For each $k$, the function $f$ is continuously differentiable on $[k,k+1]$. The one-dimensional fundamental theorem of calculus applies and gives \begin{align*} \int_k^{k+1} f'(t)\,d\mathcal L^1(t)=f(k+1)-f(k). \end{align*} Substituting this identity into the finite sum yields \begin{align*} \int_1^N A(t)f'(t)\,d\mathcal L^1(t) = \sum_{k=1}^{N-1} A_k\bigl(f(k+1)-f(k)\bigr). \end{align*}[/guided]

[step:Substitute the jump relation and telescope the finite sum] For every integer $k$ with $2\le k\le N$, the partial sums satisfy \begin{align*} a_k=A_k-A_{k-1}. \end{align*} Using the expression from the previous step, we compute \begin{align*} A_Nf(N)-\int_1^N A(t)f'(t)\,d\mathcal L^1(t) = A_Nf(N)-\sum_{k=1}^{N-1}A_k\bigl(f(k+1)-f(k)\bigr). \end{align*} Expanding the right-hand side gives \begin{align*} A_Nf(N)-\sum_{k=1}^{N-1}A_k f(k+1)+\sum_{k=1}^{N-1}A_k f(k). \end{align*} The terms with $f(j)$ for $2\le j\le N-1$ have coefficient $A_j-A_{j-1}$, the term with $f(1)$ has coefficient $A_1$, and the term with $f(N)$ has coefficient $A_N-A_{N-1}$. Hence \begin{align*} A_Nf(N)-\int_1^N A(t)f'(t)\,d\mathcal L^1(t) = A_1f(1)+\sum_{j=2}^{N-1}(A_j-A_{j-1})f(j)+(A_N-A_{N-1})f(N). \end{align*} Using $A_1=a_1$ and $A_j-A_{j-1}=a_j$ for $2\le j\le N$, this becomes \begin{align*} A_Nf(N)-\int_1^N A(t)f'(t)\,d\mathcal L^1(t) = \sum_{j=1}^N a_j f(j). \end{align*} This is exactly the desired partial summation identity. [/step]