If $f$ is continuous on $[a,b]$ and $g$ equals $f$ except at $x=c \in (a,b)$ where $g$ is defined arbitrarily. Then prove that $g$ is integrable on $[a,b]$ and $\int_{a}^{b} f = \int_{a}^{b} g$.
Here's my attempt:
I will be using this to prove it:
Let $f$ be bounded on $[a,b]$ and continuous everywhere except at one point $c$ in $[a,b]$. Then $f$ is integrable on $[a,b]$.
Clearly, if $f$ is continuous on $[a,b]$, then there is some $M\in\mathbb{R}^{+}$ such that $|f(x)| \le M$. Hence, it must be that $|g(x)| \le \max \{ M , g(c) \} $. Hence, $g(x)$ is bounded above and is continuous on $[a,b]$ except at $x=c$. Thus, by the theorem above, it is integrable.
Now, we need to show that $\int_{a}^{b} f =\int_{a}^{b} g$. If we assume that $f(c) < g(c)$ (the other case may be handled similarly), then $f(x) \le g(x)$ for all $x \in [a,b]$ and so, $\int_{a}^{b} f \le \int_{a}^{b} g$.
Let $\varepsilon >0$ be given. Then there exists a partition $P$ of $[a,b]$ such that $\int_{a}^{b} g - \varepsilon < L(P,g)$. Now we claim that for any partition $P$ of $[a,b]$, we have $L(P,g)=L(P,f)$ (we prove this claim later). Hence it follows that $\int_{a}^{b} g - \varepsilon < L(P,g) = L(P,f) \le \int_{a}^{b} f$. Since, $\varepsilon$ was arbitrary, we have $\int_{a}^{b} g \le \int_{a}^{b} f$. Hence, $\int_{a}^{b} f = \int_{a}^{b} g$.
Proof of my claim: Let $P$ be any partition of $[a,b]$. We need to show that $L(P,g) = L(P,f)$. First, we note that there must be some $i$ such that $c\in[x_{i-1} , x_{i}]$. Also, $f(x) \le g(x)$ on $[x_{i-1} , x_{i}]$, hence, $\inf_{[x_{i-1} , x_{i}]} f(x) \le \inf_{[x_{i-1} , x_{i}]} g(x)$. Also, since $f$ is continuous on $[x_{i-1} , x_{i}]$, there must be some $y\in[x_{i-1} , x_{i}]$ such that $f(y) =\inf_{[x_{i-1} , x_{i}]} f(x)$ by the Extreme Value Theorem. If $y \ne c$ then it is clear that $f(y) = g(y) =\inf_{[x_{i-1} , x_{i}]} g(x)$ otherwise if $y=c$, then for any $\varepsilon >0$, $\inf_{[x_{i-1} , x_{i}]} g(x) \le g(z)=f(z) < f(c)+\varepsilon$ for some $z \in [x_{i-1}, x_{i}]\setminus \{c\}$ ( for otherwise it would contradict our assumption that $f(c) < g(c)$). Since, $\varepsilon$ was arbitrary,$\inf_{[x_{i-1} , x_{i}]} g(x) \le f(c)$. Thus, $\inf_{[x_{i-1} , x_{i}]} f(x) = \inf_{[x_{i-1} , x_{i}]} g(x)$. Since, this is true for the rest of the subintervals as well, we must have that $L(P,g) = L(P,f)$.
Is my proof correct? I'm a bit unsure about the last part. Any alternative proofs?
EDIT: Found a easier way to complete the second part . Thanks to the hint provided by Kavi Rama Murthy.
Since, we've established that $g$ is integrable, consider the function $h:[a,b] \to \mathbb{R}$ given by $h(x)=f(x)-g(x)$. Then $h$ is integrable. Also, note that we're assuming here that $f(c) < g(c)$.
Let $P$ be any partition of $[a,b]$. Then $U(P,h)=0$ because $M_k(h)=0$ for any subintervals. Hence, it must be that $\int_{a}^{b} h = \overline{\int_{a}^{b}} h = \inf \{ U(P,h) \} =0$. Thus, again by linearity, we have $\int_{a}^{b} f = \int_{a}^{b}g$.
from Hot Weekly Questions - Mathematics Stack Exchange
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