To prove that the function $f$ is not Riemann integrable on $[a, b]$, we first recall the Riemann integrability criterion: A function is Riemann integrable on an interval $[a, b]$ if and only if it is bounded and its set of discontinuities has Lebesgue measure zero (i.e., the set of discontinuities is "small").

Step 1: Understanding the function $f(x)$

We are given that the function $f(x)$ is defined as:

• $f(x) = -1$ for all irrational $x$
• $f(x) = 1$ for all rational $x$

So, $f$ takes the value 1 for rational numbers and -1 for irrational numbers. The rationals and irrationals are dense in $\mathbb{R}$, meaning that between any two real numbers, there are both rational and irrational numbers.

Step 2: Discontinuities of $f$

To check the continuity of $f$, we need to examine the points where $f$ may fail to be continuous. A function is continuous at a point $c$ if the following holds: $\lim_{x \to c} f(x) = f(c).$ However, in our case, for any point $c \in [a, b]$, the function $f$ takes different values for rational and irrational numbers arbitrarily close to $c$:

• If $c$ is rational, $f(c) = 1$, but there are irrational numbers arbitrarily close to $c$ where $f(x) = -1$.
• Similarly, if $c$ is irrational, $f(c) = -1$, but there are rational numbers arbitrarily close to $c$ where $f(x) = 1$.

Thus, $f$ has a discontinuity at every point in $[a, b]$, because the limit does not exist due to the jump between 1 (for rationals) and -1 (for irrationals). In fact, the set of discontinuities is the entire interval $[a, b]$, which has Lebesgue measure equal to the length of the interval $(b - a)$.

Step 3: Applying the Riemann Integrability Criterion

Since the function $f$ is discontinuous at every point in $[a, b]$, the set of discontinuities has non-zero Lebesgue measure. Therefore, $f$ does not satisfy the condition of having discontinuities of measure zero, which is required for Riemann integrability. Hence, $f$ is not Riemann integrable on $[a, b]$.

Step 4: What about $f^2$?

Now, let’s consider $f^2$. Since $f(x) = 1$ for rational $x$ and $f(x) = -1$ for irrational $x$, we have: $f^2(x) = 1^2 = 1 \quad \text{for all} \, x \in \mathbb{R}.$ Therefore, $f^2(x) = 1$ for every $x$, both rational and irrational. This function is constant, and constant functions are always continuous and Riemann integrable on any interval $[a, b]$.

Thus, $f^2$ is Riemann integrable on $[a, b]$, and its integral is simply the area under the curve, which is the length of the interval: $\int_a^b f^2(x) \, dx = \int_a^b 1 \, dx = b - a.$

Conclusion:

• The function $f(x)$ is not Riemann integrable on $[a, b]$.
• The function $f^2(x)$ is Riemann integrable, and its integral on $[a, b]$ is $b - a$.

Would you like further details on any of these steps, or is there anything else you’d like to explore? Here are some related questions:

1. Why are the rationals and irrationals dense in $\mathbb{R}$?
2. How does the concept of Lebesgue measure relate to Riemann integrability?
3. Can a function with a finite number of discontinuities be Riemann integrable?
4. What are other common tests for determining if a function is Riemann integrable?
5. Can you prove that any function with a jump discontinuity at every point is not Riemann integrable?

Tip: A useful approach to checking Riemann integrability is to consider the set of discontinuities and apply the criterion that the function is integrable if this set has measure zero.