# What is wrong with my contradiction?

Spivak says the following function does not have an integral:

$$F(x) = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right.$$

It makes sense, for any partition $P$ there will be a $t_i$ and $t_{i+1}$ where there will exist a rational and an irrational between the two, hence $M_{i}$ and $m_i$ (the "supremum partition" and "infimum partition") are always going to to be 1 and 0 respectively. Hence if you take the sum of these different partitions and subtract them from each other (as Spivak writes: $U(F, P) - L(F, P)$) you will always get a $b-a$ where $b$ and $a$ are the upper and lower bounds of the integral respectively.

Therefore, I can define a $\epsilon < a - b$ and I will never be able to make a partition that results in $U(F, P) - L(F, P) < \epsilon$ hence it fails.

It makes sense, but my alternative reasoning comes to the complete opposite conclusion. First of all, Spivak proves the following two properties:

1. Let $f$ be a function where for every point $x$, $f(x) = 0$ except for exactly one point $a$ where $f(a) = 1$ then $f$ is integrable. Let this kind of function be referred to as a "point function".

2. Let $f$ and $g$ be 2 integrable functions, then $f + g$ is integrable.

After that I decide to generalize the second theorem:

Lemma:

Let $f, g, h \ldots n$ all be integrable function with signatures of $\mathbb{R} \rightarrow \mathbb{R}$, then $f + g + h + \ldots + n$ is integrable.

Proof:

Base case:

$f + g$ is integrable by the theorem above, now assume it holds true for $f + g + \ldots + m$ then to show it is true, it must work for $f + g + \ldots + m + n$.

$$f + g + \ldots + m + n$$

Let the integrable function $j$ denote $f + g + \ldots + m$ then we have:

$$j + n$$

Since $j$ and $n$ are both integrable functions, by Theorem 2 we have shown this works for an arbitrary amount of functions.

Now, let $A$ be the set of all point functions with a rational number. Then: $\sum{A}$ is integrable since we know that all the functions in $A$ are integrable and by the above lemma the some of them all must result in an integrable function. And since:

$$\sum{A} = F$$

I have shown $F$ is integrable, which it is clearly not. (Also, using my logic you can argue the same for any arbitrary function.) Can someone help me understand where I went wrong?

• You have the theorem that the sum of an arbitrary finite number of integrable functions is integrable. You can't conclude that the sum of infinitely many integrable functions is integrable. – Daniel Fischer Apr 16 '14 at 20:24
• @Daniel Fischer: Hmmm, didn't know that induction only works on finite sets... That would cause this to fail if that isn't true... Thought the whole point of induction was to be able to do stuff with infinite. Well, that was rather a silly mistake on my part. – Dair Apr 16 '14 at 20:27
• The very example you are studying is one illustration why the additive property of integrals fails for infinite decompositions. Here, in some sense $F(x) = \sum_{q \in \mathbb{Q}} f_q(x)$ where $f_q(x) = 1$ if $x = q$ and $= 0$ otherwise. (The 'in some sense' comment is that you need to define what you mean by an infinite sum of functions, but this can be done rigorously.) Then $\int_a^b \sum_{q \in \mathbb{Q}} f_q(x) \, dx$ does not exist, even though $\sum_{q \in \mathbb{Q}} \int_a^b f_q(x) \, dx = 0$ does. +1 btw for explaining your thought process well in your question. – Michael Joyce Apr 16 '14 at 20:45
• A way to think about induction is that it proves infinitely many finite cases. Induction proves statements of the form "For all $n \in \mathbb{N}$, Proposition $P_n$ is true." For example, it proves that an integral of any finite sum of $n$ integrable functions is equal to the finite sum of the integrals of each function; there are infinitely many statements proven (one for each $n$), but not one of the statements involves an infinite sum. – Michael Joyce Apr 16 '14 at 21:02
• You have asked a very interesting question that leads to the core of some important questions in real analysis, about infinity and infinite sums and which functions are integrable. Congratulations (and congratulations for reading Spivak on your own). Why not use the comments to answer your own question here, so others can learn what you're learning. – Ethan Bolker Jun 27 '14 at 15:58

Therefore the Lemma shown above does not apply to $A$.