Is the Function Definition with 9^n Well-Defined?

The displayed rule assigns a value to rational inputs, but its irrational case contains an undefined symbol, $n$. Unless $n$ is a fixed number or a specified function of $x$, the formula does not define one particular function. A simple interpretation is that $n$ is a fixed integer; under that interpretation, the function is measurable and its Lebesgue integral can be calculated directly.

If $n$ is fixed

Let $n$ be a fixed integer and define

$$f(x)= \begin{cases} 0, & x\in\mathbb{Q},\\ 9^n, & x\notin\mathbb{Q}, \end{cases} \qquad 0\leq x\leq 1.$$

This is a two-valued function. The rationals in $[0,1]$ form a countable set, so they have Lebesgue measure zero. The irrationals make up the rest of the interval and have measure one. Thus $f$ equals the constant $9^n$ almost everywhere, meaning everywhere except on a set of measure zero.

The function is Lebesgue measurable: the sets on which it takes its two possible values are the rationals and the irrationals, both measurable. Its Lebesgue integral is

$$\int_0^1 f(x)\,dx =0\cdot m(\mathbb{Q}\cap[0,1])+9^n\cdot m([0,1]\setminus\mathbb{Q}) =9^n,$$

where $m$ denotes Lebesgue measure. This holds for any fixed integer $n$; for example, if $n=2$, then $f$ is $81$ on the irrationals and its integral is $81$.

Worked example when the exponent varies

If the intended meaning is that $n$ depends on $x$, its rule must be given. For one possible rule, set $n(x)=1$ on irrational numbers in $[0,1/2)$ and $n(x)=2$ on irrational numbers in $[1/2,1]$, while keeping $f(x)=0$ for rational $x$. The resulting function is measurable, and its integral is

$$\int_0^1 f(x)\,dx =9\,m([0,1/2)\setminus\mathbb{Q}) +81\,m([1/2,1]\setminus\mathbb{Q}) =9\cdot\frac12+81\cdot\frac12 =45.$$

The rationals at the endpoints or dividing point do not change the integral, because a finite set has measure zero. A different assignment of exponents to irrational inputs can give a different integral, or even make the function nonmeasurable.

More generally, if $n(x)$ takes integer values, define

$$A_k=\{x\in[0,1]\setminus\mathbb{Q}:n(x)=k\}.$$

When all these level sets are measurable, the function is measurable and its nonnegative Lebesgue integral is

$$\int_0^1 f\,dm=\sum_k 9^k m(A_k),$$

with the sum allowed to be infinite. Without the rule for $n(x)$, neither the level sets nor the integral can be determined.

Lebesgue versus Riemann integrability

For fixed $n$, the function is bounded and Lebesgue integrable. It is not Riemann integrable, because $9^n\ne0$ and both the rationals and irrationals are dense in $[0,1]$. At every point, values arbitrarily close to that point include both $0$ and $9^n$, so the function is discontinuous everywhere. A bounded function on a closed interval is Riemann integrable only if its set of discontinuities has measure zero; here that set is the entire interval.

Common mistakes

• Treating an unspecified $n$ as if it were fixed. A fixed exponent and an exponent that depends on $x$ describe different functions.
• Confusing measure with cardinality. The rationals are infinite, but countable; their Lebesgue measure is zero.
• Concluding nonmeasurability from discontinuity. A function can be discontinuous everywhere and still be Lebesgue measurable, as in the fixed-$n$ case.
• Assuming every variable-exponent rule works. If $n$ varies, its level sets must be specified; arbitrary assignments need not be measurable.