Integrability after removing a point (integrating and discontinuity)

Question

Below is a problem that I am trying to solve.

Assume $f:[a,b] \to \mathbb{R}$ is integrable.

(a) Show that if one value of $f(x)$ is changed at some point $x \in > [a,b]$, then $f$ is still integrable and integrates to the same value as before

(b) Show that the observation in (a) holds if a finite number of values of $f$ are changed.

(c) Find an example to show that by altering a countable number of values, $f$ may fail to be integrable.

I began solving part (a). What I have so far:

(a) Assume $f$ is integrable so that $U(f)=L(f)$. Now, let $g$ be the modified function where we have changed the value of $f$ at some $x_{0} \in [a,b]$. We claim that $U(g)=U(f)$ and $L(g)=L(f)$.

Let $\epsilon >0$. To show that $U(g)=U(f)$, we want to find a partition $P$ for which $U(g,P)< U(f)+ \epsilon$.

I am not sure if I started on the right track, or how to continue explaining (a). Hopefully once I figure out (a), I can complete the rest. Any hints would be extremely helpful - thank you!

Edit: There is a similar problem that was also posted (Integrability doesn't change if we remove one point), but most of the solutions appear to be building off the submitted suggestion for a proof, which I am not sure I understand. My main issue is I am having a hard time finding what I am being asked to prove. I believe I would benefit more from an explanation as to why I am not on the right track, or perhaps a hint as to what I should look at first when trying to approach the proof.

Edit 2: I read this question and found it somewhat useful. It was relatively easy to follow.

Answer 1

What the question asks is (i) start with a Riemann integrable function $f$ (ii) consider another function $g$ [not known to be integrable] (iii) assume that $f=g$ except at finitely many points. (iv) Now show that $g$ is integrable and the two integrals have the same value.

Bad advice: try to use the definition of the integral for $f$ and $g$ and mess around with Riemann sums or Darboux sums. Try real hard to show $g$ is integrable without getting lost in all those inequalities, etc.

Good advice: Note that $g = f -h$ where $h=f-g$ is a very simple function [it is zero except at a few points]. If both $f$ and $h$ are integrable so is $g$ etc. Bright idea: focus your attention on $h$.

To make it clear so that you can see the logic in the steps:

1. Show that if $h:[a,b]\to \mathbb R$ is zero everywhere but with one exception, then $h$ is Riemann integrable and $\int_a^b h(x)\,dx=0$.

Hint: get into the details of Riemann or Darboux sums for this.

2. By induction (or otherwise) show that if a function $h:[a,b]\to \mathbb R$ is zero everywhere but with finitely many exception, then $h$ is Riemann integrable and $\int_a^b f(x)\,dx=0$.

3. Recall the fact that $f$ and $h$ are both Riemann integrable so too is the sum $f+h$ and the difference $f-h$.

4. Conclude that if $f$ is Riemann integrable and $g$ differs from $f$ at a finite number of points, then $g$ is also Riemann integrable and, moreover, $\int_a^b f(x)\,dx= \int_a^b g(x)\,dx$.

Hint: Write $h(x)=f(x)-g(x)$ and apply #2 and #3.

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Here is a follow-up that is more general, more interesting, but uses mostly the same ideas;

Theorem. Suppose that $ f $ and $ g $ are bounded functions on an interval $ [a,b] $ and that they differ at only a finite number of points. Then $$ \underline{\int_a^b} f(x)\,dx = \underline{\int_a^b} g(x)\,dx. $$ and $$ \overline{\int_a^b} f(x)\,dx = \overline{\int_a^b} g(x)\,dx. $$

Then your problem is an immediate corollary.

Never be satisfied with some messy proof with $ \epsilon $s and sups/infs. Try to write up in transparent steps.

Hint: If $ f $ and $ g $ are bounded functions on an interval $ [a,b] $ and they differ at only a finite number of points, then take a good look at the function $ h = f-g. $ What can you say and how might that help?

Definition. A function $ h $ on an interval $ [a,b] $ will be called trivial if it is zero except possibly at a finite number of points.

Lemma 1. If $ h $ is a trivial function on an interval $ [a,b] $ then $ h $ is Riemann integrable there and $ \int_a^b h(x)\,dx =0 $.

Not hard to prove. This was step #1 and #2 in the hints above.

Lemma 2. If $ f $ and $ g $ are bounded functions on an interval $ [a,b] $ then $ \underline{\int_a^b} f(x)\,dx + \underline{\int_a^b} g(x)\,dx \leq \underline{\int_a^b} [ f(x)+g(x)]\,dx $

This is standard and should be in any text or lecture that discusses such things. Important to know anyway. Note the inequality. Not an equality.

Lemma 3. Suppose that $ f $ and $ g $ are bounded functions on an interval $ [a,b] $ and that they differ at only a finite number of points. Then $ \underline{\int_a^b} f(x)\,dx = \underline{\int_a^b} g(x)\,dx. $ A similar statement is true for the upper integrals.

Proof. Let $ h = f-g $ . Then $ h $ is trivial. Apply Lemma 1 and Lemma 2 to obtain $$ \underline{\int_a^b} g(x)\,dx = \underline{\int_a^b} g(x)\,dx + \underline{\int_a^b} h(x)\,dx $$ $$ \leq \underline{\int_a^b} [ g(x)+h(x)]\,dx = \underline{\int_a^b} f(x)\,dx .$$ Similarly $$ \underline{\int_a^b} f(x)\,dx = \underline{\int_a^b} f(x)\,dx + \underline{\int_a^b} [- h(x)]\,dx $$ $$ \leq \underline{\int_a^b} [ f(x)+[-h(x)]]\,dx = \underline{\int_a^b} g(x)\,dx. $$

QED.

Note too this usual trick. You want to prove that $ A = B $? Well show instead that $ A\leq B $ and also that $ B\leq A $. Inequalities are very often easier than equalities.

Answer 2

Objectively speaking, you are on a right track since any proof of the theorem will imply that $U(f) = U(g)$, and conversely, once you prove $U(f) = U(g)$ you are done proving the theorem since the argument for $L(f) = L(g)$ is the same.

Since $f$ and $g$ only disagree at $x_0$, one statement that is easy to prove is that for any partition $P$ of $[a, b]$ and $\varepsilon > 0$, there is a refinement $P'$ such that $|U(f, P') - U(g, P')| \leq \varepsilon$. To prove it, just refine $P$ so that the subintervals containing $x_0$ have length $\leq \varepsilon$.

Now once you have this, you are done since by definition of $U$, there is $P$ such that $$U(f) \leq U(f, P) < U(f) + \varepsilon$$ and by refining $P$, we can assume that $$U(g) \leq U(g, P) < U(g) + \varepsilon,$$ $$|U(g, P) - U(f, P)| < \varepsilon$$ as well.

Note the similarity in this proof to the proof of linearity of integration, which, as the other answer alluded to, can be used to prove this theorem as well.