This question is a kind of follow-up of this question, since I realized in subsequent refinement of the contents of the proof that the proof I presented here does actually not hold. For ease of reference, I will restate the problem at hand, here.
Suppose $f:[a,b]\to\mathbb{R}$ and $g:[a,b]\to\mathbb{R}$ are bounded functions such that $f(x)=g(x)$, for each $x$ in $[a,b]$, apart from a finite number of points. Then, using the fact that, if $\langle P_n\rangle$ is a limiting sequence of partitions of $[a,b]$, then $$ \lim_{n\to\infty}\mathcal{L}(f,P_n)=\underline{\int_a^b}f(x)~dx $$ Where $\mathcal{L}(f,P_n)$ denotes the lower Darboux sum of $f$ on the interval $[a,b]$, with respect to the partition $P_n$, prove that $$ \underline{\int_a^b}f(x)~dx=\underline{\int_a^b}g(x)~dx $$
This is my latest attempt.
Let $\gamma=\lbrace\gamma_1,\gamma_2,\cdots,\gamma_k\rbrace\subset[a,b]$ denote the set of values of $x$ such that $f(x)\neq g(x)$, and $\gamma_1<\gamma_2<\cdots<\gamma_k$. Now define $$ \Gamma=\min_{1\leq i<k}(\gamma_{i+1}-\gamma_i) $$ And hence let $\langle P_n\rangle$ be a limiting sequence of partitions of $[a,b]$ such that $|P_n|=k_n$. Therefore we know that $\exists N\in\mathbb{N}$ such that $n>N\implies ||P_n||<\Gamma$. Hence, let $n>N$, and define $I=\lbrace i_j:1\leq j\leq k\rbrace$ to be the set of values such that $\gamma_j\in[x_{i_j-1},x_{i_j}]$, for some $x_{i_j-1},x_{i_j}\in P_n$. Now let $$ m_i=\inf_{x\in[x_{i-1},x_i]}f(x),~~~\mathrm{and}~~~w_i=\inf_{x\in[x_{i-1},x_i]}g(x),~~~x_{i-1},x_i\in P_n $$ Then we have that $$ |\mathcal{L}(f,P_n)-\mathcal{L}(g,P_n)|=\left|\sum_{i=1}^{k_n}m_i(x_i-x_{i-1})-\sum_{i=1}^{k_n}w_i(x_i-x_{i-1})\right| $$ $$ =\left|\sum_{i=1}^{k_n}(m_i(x_i-x_{i-1})-w_i(x_i-x_{i-1}))\right| $$ $$ =\left|\sum_{i=1}^{k_n}(m_i-w_i)(x_i-x_{i-1})\right| $$ $$ =\left|\sum_{i\notin I}(m_i-w_i)(x_i-x_{i-1})+\sum_{i\in I}(m_i-w_i)(x_i-x_{i-1})\right| $$ $$ =\left|\sum_{i\notin I}0(x_i-x_{i-1})+\sum_{i\in I}(m_i-w_i)(x_i-x_{i-1})\right| $$ $$ =\left|\sum_{i\in I}(m_i-w_i)(x_i-x_{i-1})\right| $$ $$ \leq||P_n||\left|\sum_{i\in I}(m_i-w_i)\right| $$ Now, since $I$ is a finite set (in particular, $|I|=k$), and both $f$ and $g$ are bounded, meaning $m_i-w_i$ is necessarily finite $\forall i\in I$, we have that $\exists M\in\mathbb{R}$ satisfying $$ \left|\sum_{i\in I}(m_i-w_i)\right|\leq M $$ Hence, $\forall n$ in some neighbourhood of $+\infty$, we have that $0\leq|\mathcal{L}(f,P_n)-\mathcal{L}(g,P_n)|\leq M||P_n||$, and thus, we may apply the sandwich theorem to arrive at the conclusion that $|\mathcal{L}(f,P_n)-\mathcal{L}(g,P_n)|\to0$ as $n\to\infty$. Hence $\mathcal{L}(f,P_n)-\mathcal{L}(g,P_n)\to0$ as $n\to\infty$, and thus observe that $$ \lim_{n\to\infty}(\mathcal{L}(f,P_n)-\mathcal{L}(g,P_n))=0 $$ $$ \lim_{n\to\infty}\mathcal{L}(f,P_n)-\lim_{n\to\infty}\mathcal{L}(g,P_n)=0 $$ $$ \underline{\int_a^b}f(x)~dx-\underline{\int_a^b}g(x)~dx=0 $$ $$ \therefore\underline{\int_a^b}f(x)~dx=\underline{\int_a^b}g(x)~dx $$ As required. This completes our proof.
Again, I'm not only asking whether or not the proof is correct, I'm more concerned with stylistic tips, or whether or not there is perhaps a more elegant variation of this proof that I happen to be missing. Any help is appreciated, thank you.