Question

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.

Answer 1

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 \\`.