Proving that the integrals of two functions are the same if they are equal everywhere except a point

Question

Let $f(x)$ and $g(x)$ be integrable functions over $[a,b]$ and let $∂$ be a point on $[a,b]$. If $f(x) = g(x)$ for all $x≠∂$, then $$\int_a^b f(x)dx=\int_a^b g(x)dx$$

Answer 1

We can look at this from another angle. What we need to show is that $$\int_a^b (f(x)-g(x)) dx = 0.$$ The function $h(x) = f(x)-g(x)$ is zero except at the one point that I will relabel as $p$.

Now suppose that $\delta >0$ and small enough so that $(p-\delta, p+\delta) \subset (a,b)$.

Then $$\int_a^b (f(x) - g(x)) dx = \int_{p-\delta}^{p+\delta} (f(x) - g(x)) dx$$ since $f(x)-g(x)$ is zero outside of $(p-\delta, p+\delta)$.

Moreover: $$\left| \int_{p-\delta}^{p+\delta} (f(x) - g(x)) dx\right| \le \int_{p-\delta}^{p+\delta} |f(x) - g(x)| dx \le \int_{p-\delta}^{p+\delta} |f(p)-g(p)| dx = 2\delta |f(p)-g(p)|.$$

Now as we let $\delta$ get very small, the quantity on the right tends to zero. Thus $$\int_a^b (f(x) - g(x)) dx = \int_{p-\delta}^{p+\delta} (f(x) - g(x)) dx \to 0$$ as $\delta \to 0$, and we conclude $$\int_a^b (f(x) - g(x)) dx =0.$$