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Assume f : [a, b] Ris integrable. Show that if g satisfies g(x)=f(x) for all but a finite number of points in (a, b), then g is integrable as well and S8= Suf. a Hint: start by considering one x where f and g differ.

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Assume
\sf\:f : [a, b] \rightarrow \mathbb{R} \\ is integrable. We want to show that if
\sf\:g satisfies
\sf\:g(x) = f(x) \\ for all but a finite number of points in
\sf\:(a, b), then
\sf\:g is integrable as well and
\sf\:\int_(a)^(b) g(x) \, dx = \int_(a)^(b) f(x) \, dx \\.

Proof:

Assume there exists a point
\sf\:x_0 in
\sf\:(a, b) where
\sf\:f and
\sf\:g differ.

Since
\sf\:f is integrable, by definition, for any given
\sf\:\epsilon > 0 \\, there exists a partition
\sf\:P of
\sf\:[a, b] such that the upper sum
\sf\:U(f, P) - L(f, P) < \epsilon \\.

Let's consider the partition
\sf\:P which includes the point
\sf\:x_0. Without loss of generality, assume
\sf\:x_0 is the right endpoint of a subinterval
\sf\:[x_i, x_(i+1)] \\ in
\sf\:P.

Since
\sf\:g(x) = f(x) \\ for all but a finite number of points, we can select a subinterval
\sf\:[x_i, x_(i+1)] \\ such that
\sf\:g(x) = f(x) \\ for all
\sf\:x in
\sf\:[x_i, x_(i+1)] \\except
\sf\:x_0.

Now, we have:


\sf\:\sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) = \sup_{x \in [x_i, x_(i+1)]} f(x) - \inf_{x \in [x_i, x_(i+1)]} f(x) \\

Using this, we can rewrite the upper sum and lower sum as:


\sf\:U(g, P) = U(f, P) + \sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) \\


\sf\:L(g, P) = L(f, P) + \sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) \\

Since
f is integrable, we have
\sf\:U(f, P) - L(f, P) < \epsilon \\. Therefore,
\sf\:U(g, P) - L(g, P) < \epsilon \\ as well.

This holds for any partition
\sf\:P of
\sf\:[a, b]. Hence,
\sf\:g is integrable and
\sf\:\int_(a)^(b) g(x) \, dx = \int_(a)^(b) f(x) \, dx \\.

User Zeno Tsang
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