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Let $f: [a,b] \rightarrow \mathbb{R}$ and $g: [a,b] \rightarrow \mathbb{R}$ be bounded. Suppose $S = \{x \in [a,b] : f(x) \neq g(x)\}$ is finite. Prove that if $f$ is Riemann integrable on $[a,b]$ then so is $g$ and $\int^b_a f = \int^b_a g$.

This is a homework problem that I have no idea how to do.

Professor gave the following hint: Let $h = g - f$. Then by a theorem in the book, if $h$ is Riemann integrable, then so is $g = h + f$. So it suffices to prove any function on $[a,b]$ which is zero except at a finite number of points is Riemann integrable. By induction, you will only need to show a function on $[a,b]$ which is zero except at exactly one point is Riemann integrable.

Thanks for any help.

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    $\begingroup$ So what's the problem with proving that a function with a zero except at a single point is riemann integrable, and has integral zero? If you start from the definition of the riemann integral, that shouldn't be too hard... $\endgroup$ – fgp May 4 '14 at 0:35
  • $\begingroup$ Prove that if $f$ is bounded on $[\alpha,\beta]$ and integrable on $[\alpha,\gamma]$ for all $\gamma\in(\alpha,\beta)$, then $f$ is integrable in $[\alpha,\beta]$. Same if $f$ is assumed integrable in $[\gamma,\beta]$ for all such $\gamma$. Use this to verify the result you need. $\endgroup$ – Andrés E. Caicedo May 4 '14 at 0:35
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    $\begingroup$ (My suggestion actually gives you a way of proving that if $f$ is bounded on $[a,b]$, and continuous except at finitely many points, then it is integrable on $[a,b]$.) $\endgroup$ – Andrés E. Caicedo May 4 '14 at 0:36
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A function is Riemann integrable on $[a, b]$ if $\sup L(P, f) = \inf U(P, f)$ where the supremum and infimum are taken over finite partitions of $[a, b]$.

Let $z \in [a, b]$. Suppose $h: [a, b] \to \mathbb{R}$ with $h(z) > 0$ and $h(x) = 0$ for all $x \in [a, b]$ with $x \neq z$.

$L(P, f) = 0$ because the interval that includes $z$ also includes other points, which $h$ maps to zero.

For any $P$, we have $0 = L(P, f) < U(P, f)$. So zero is a lower bound for the set of values taken by $U$. Let $\Delta_z$ be the length of the interval that contains $z$. Then $U(P, f) = h(z) \Delta_z$. Thus $U(P, f)$ can be made smaller than any positive number by choosing a partition with $\Delta_z$ sufficiently small. So $\inf U(P, f) = 0$.

Thus $h$ is Riemann integrable and $\int_a^b h(x) dx = 0$.

You should be be able to do the case where $h(z) < 0$.

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