Question

Prove a function that differs in finitely many points has the same integral as another

Answer 1

I'll show a proof if the two functions differ for only one point. Then, you can repeat the same argument in the neighbourhood of every other point.

So, suppose that we have

$f:[a,b]\to R,\quad g:[a,b]\to R,\quad \begin{cases} f(x) = g(x) &\text{ if } x\neq x_0\\ f(x_0)\neq g(x_0) \end{cases}$

We can write

$\displaystyle \int_a^b f(x)\; dx = \int_a^{x_0-\varepsilon} f(x)\; dx + \int_{x_0-\varepsilon}^{x_0+\varepsilon} f(x)\; dx + \int_{x_0-\varepsilon}^b f(x)\; dx$

We can write the exact same thing for $g(x)$, and we have (as far as we know)

$\displaystyle \int_a^{x_0-\varepsilon} f(x)\; dx = \int_a^{x_0-\varepsilon} g(x)\; dx,\ \int_{x_0-\varepsilon}^{x_0+\varepsilon} f(x) \neq \int_{x_0-\varepsilon}^{x_0+\varepsilon} g(x)\; dx,\ \int_{x_0-\varepsilon}^b f(x)\; dx = \int_{x_0-\varepsilon}^b g(x)\; dx$

So, we only need to prove that the term where f and g are not the same function can be made arbitrarily small.

Since f and g are integrable on an interval, they are bounded. Let's call $M_f$ and $M_g$ the maximum of f and g, respectively, in the interval $[x_0-\varepsilon, x_0 + \varepsilon]$. We have

$\displaystyle \int_{x_0-\varepsilon}^{x_0 + \varepsilon} f(x)\; dx \leq \int_{x_0-\varepsilon}^{x_0 + \varepsilon} M_f\; dx = 2M_f\varepsilon$

The same goes for g(x). So, You have

$\displaystyle \left|\int_a^b f(x)\;dx - \int_a^b g(x)\;dx\right| \leq 2(M_f+M_g)\varepsilon$

which tends to 0 as $\varepsilon\to 0$