$f: [0,1] \to \mathbb{R}$ such that $f(x)=1$ if $x=\frac{1}{2},\frac{1}{4},...$ and $f(x)=0$ else is integrable

Question

Theorem statement:

The function $f: [0,1] \to \mathbb{R}$ given by $f(x)=1$ for $x=\frac{1}{2},\frac{1}{4},\frac{1}{8},...$ and $f(x)=0$ otherwise is integrable.

My proof:

Partition $[0,1]$ as follows $ 0 \leq ... \leq \frac{1}{2^{k}} \leq ... \leq \frac{1}{4} \leq \frac{1}{2} \leq 1$. Then for minorant step functions $ \phi_{-} \leq f$ we have $\phi_{-} \leq 0$ on $(\frac{1}{2^{j}}, \frac{1}{2^{j-1}}$) and so $\sup I(\phi_{-}) =0$. Similarly we can argue for majorant step functions $\phi_{+} \leq f $ that $\inf I(\phi_{-}) =0$ on $(\frac{1}{2^{j}}, \frac{1}{2^{j-1}}$) . We conclude that $\inf I(\phi_{-}) = \sup I(\phi_{-}) =0$, so f is indeed Riemann integrable on $[0,1]$.

I have a feeling there is something seriously wrong with this argument. If any one would tell me where (if at all) my argument falls apart, I would be very grateful.

Answer 1

Partition $[0,1]$ as follows $$0 \leq ... \leq \frac{1}{2^{k}} \leq ... \leq \frac{1}{4} \leq \frac{1}{2} \leq 1.$$

Note that in Riemann and/or Darboux sums, each partition has finitely many partition points.

We conclude that $\inf I(\phi_{-}) = \sup I(\phi_{-}) =0$.

In Darboux sums, one has to take sup/inf from closed subintervals formed by adjacent partition points. Therefore, your argument is flawed.

There's no denying that the function values $f(a), f(b)$ at the endpoints won't affect the value of the definite integral $\int_a^b f$ provided that $f$ is integrable on $[a,b]$. To see this, suppose that $f$ is bounded on $[a,b]$ by $M > 0$. Changing $f(a)$ and $f(b)$ only affects the blue area (, which is $4M\epsilon$), which can be set to arbitrarily small.

Now, to fix your proof, you only need to choose finitely many appropriate blue strips with arbitrarily small area to cover the jump discontinuities. Since those discontinuities are "concentrated" at $x = 0$, you don't need the blue strip at the right endpoint $x = 1$. You'll only need one blue strip at the left endpoint $x = 0$. That will cover the discontinuities except at $x = 2^j$, where $j = 1, \dots, \lfloor -\log_2\varepsilon \rfloor, \lceil -\log_2\varepsilon \rceil$. The last two candidates of $j$ might get covered by the left blue strip, but I'm adding them as well. Cover these discontinuities too with non-overlapping blue strips with thinner width, so that the total area of blue strips is still arbitrarily small. This is possible since there're only finitely many possible $j$, and the range of $j$ only depend on $\varepsilon$.

Answer 2

You can prove the following results step by step:

Theorem: Let $f:[a, b] \to\mathbb{R} $ be a bounded function and let $D$ be the set of its discontinuities.

• If $D=\emptyset $ then $f$ is Riemann integrable on $[a, b] $.
• If $D$ is finite then $f$ is Riemann integrable on $[a, b] $.
• If $D$ has a finite number of limit/accumulation points in $[a, b] $ then $f$ is Riemann integrable on $[a, b] $.

For the function in question the third criterion applies as the set of discontinuities has only one limit point $0$ and thus the function is Riemann integrable.

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Note: The theorem above does not cover the case when $D$ is countable. The result holds in this case too and can be proved without measure theory but with more difficulty.