Changing an integrable function at finitely many points changes neither its integrability nor its integral

The one-point case is proved first, for a function called $w$ so that no symbol is reused. Let $p \in [a,b]$ and let $w : [a,b] \to \mathbb{R}$ satisfy $w(x) = 0$ for every $x \ne p$; put $K := |w(p)|$, so $|w(x)| \le K$ for every $x$ and $w$ is bounded.

Fix $N \ge 1$ and write $U_N = (N,t)$ with subintervals $I_i$ and lengths $\Delta_i = (b-a)/\iota(N)$. Define $\theta_i := 1$ if $p \in I_i$ and $\theta_i := 0$ otherwise, for $i < N$.

Setting up the induction. Put $h := g - f$, so that $h(x) = 0$ for every $x \notin E$, and for $j < r$ define $h_j : [a,b] \to \mathbb{R}$ by $h_j(x) := h(x)$ if $x = e(j)$ and $h_j(x) := 0$ otherwise. Each $h_j$ vanishes off the single point $e(j)$.

At most two indices $i < N$ have $\theta_i = 1$, and they are consecutive: if $p \in I_i \cap I_j$ with $i < j$ then $t_j \le p \le t_{i+1}$ and $t_{i+1} \le t_j$, so $t_{i+1} = t_j$ and $j = i+1$ by injectivity of $t$. Also some index has $\theta_i = 1$, since the subintervals cover $[a,b]$; let $i_0$ be the least such.

For every $x \in [a,b]$ and every $k \le r$, $\sum_{j<k} h_j(x) = h(x)$ when $x = e(j_0)$ for some $j_0 < k$, and $\sum_{j<k}h_j(x) = 0$ otherwise: in the first case all terms with $j \ne j_0$ vanish, because $e$ is injective, and [L4] evaluates the sum; in the second every term is $0$.

For every $i < N$: if $\theta_i = 0$ then $w$ vanishes on $I_i$, so $m_i = M_i = 0$, where $m_i$ and $M_i$ are the extreme values of $w$ on $I_i$; and always $-K\theta_i \le m_i \le M_i \le K\theta_i$.

Define $y_i := \Delta_i$ for $i = i_0$ and $y_i := 0$ otherwise, and $z_i := \Delta_i$ for $i = i_0+1$ with $i_0 + 1 < N$, and $z_i := 0$ otherwise. Then $\theta_i\Delta_i \le y_i + z_i$ for every $i < N$ by step 2.1, and $\sum_{i<N}(y_i+z_i) \le 2(b-a)/\iota(N)$ by [L4] and [L3].

Let $Q(k)$, for $k \le r$, be the statement that the function $H_k : x \mapsto \sum_{j<k}h_j(x)$ is integrable on $[a,b]$ with $\int_a^b H_k = 0$.

By step 3.1 and monotonicity of finite sums, $U(w,U_N) - L(w,U_N) = \sum_{i<N}(M_i-m_i)\Delta_i \le \sum_{i<N}2K\theta_i\Delta_i \le 4K(b-a)/\iota(N)$, and likewise $-2K(b-a)/\iota(N) \le L(w,U_N)$ and $U(w,U_N) \le 2K(b-a)/\iota(N)$.

Base. $H_0$ is the constant function $0$, integrable with integral $0$ by [L5], so $Q(0)$ holds.

Induction hypothesis. Fix $k < r$ and assume $Q(k)$.

Given a real $\eta > 0$, [L6] supplies $N$ with $4K(b-a)/\iota(N) < \eta$, so $w$ satisfies Riemann's criterion and is integrable by [L1].

Moreover $|\int_a^b w| \le 2K(b-a)/\iota(N)$ for every $N \ge 1$ by step 4.1 and [L2], and the right-hand side is below every positive real by [L6]; hence $\int_a^b w = 0$. Steps 1.1 to 5.2 therefore prove: every function on $[a,b]$ vanishing off a single point is integrable with integral $0$.

$H_{k+1} = H_k + h_k$ pointwise by [L4], and $h_k$ is integrable with integral $0$ by steps 5.1 and 5.2 applied to $w := h_k$ and $p := e(k)$; so $H_{k+1}$ is integrable with $\int_a^b H_{k+1} = 0 + 0 = 0$ by [L5], which is $Q(k+1)$.

By [L7] with steps 4.2 and 6.1, $Q(k)$ holds for every $k \le r$; at $k = r$, and by step 2.2, $H_r = h$, so $h$ is integrable with $\int_a^b h = 0$.

Hence $g = f + h$ is integrable with $\int_a^b g = \int_a^b f + 0 = \int_a^b f$ by [L5].