📝 Discontinuities and integrability (24 MCQs)

📖 Explanation: This is the Lebesgue criterion for Riemann integrability: a bounded function is Riemann integrable if and only if its set of discontinuities has measure zero. While continuity, finite discontinuities, or monotonicity are sufficient conditions, they are not necessary. For example, the Dirichlet function is discontinuous everywhere and not integrable, but the Thomae function is discontinuous on a measure-zero set (the rationals) and is integrable despite having infinitely many discontinuities.

📖 Explanation: The Dirichlet function is the classic example of a bounded function that fails to be Riemann integrable. It is discontinuous at every real number (both rational and irrational points). Since its set of discontinuities is the entire interval $[0,1]$, which does not have measure zero, it violates the Lebesgue criterion for integrability. This function is not Riemann integrable despite being bounded because no matter how fine the partition, the Riemann sums do not converge to a single value.

📖 Explanation: The function $f(x) = 1/x$ is unbounded on $[0,1]$ because as $x$ approaches 0 from the right, $f(x)$ tends to infinity. A necessary condition for Riemann integrability is that the function must be bounded on the interval. Even though it is continuous on $(0,1]$, the vertical asymptote at $x = 0$ introduces unboundedness. The single point of discontinuity at $x = 0$ would not by itself prevent integrability if the function were bounded, but unboundedness makes the Riemann sums arbitrarily large.

📖 Explanation: The function $\sin(1/x)$ is famous for its wild oscillation near $x = 0$. However, it is bounded (between -1 and 1) and has a single point of discontinuity at $x = 0$ after defining $f(0) = 0$. A function with finitely many discontinuities on a closed interval is Riemann integrable. The infinite oscillations do not prevent integrability as long as the function is bounded; the Riemann sums converge because the oscillatory behavior is confined to a neighborhood that can be made arbitrarily small.

📖 Explanation: The student's statement is false. A function with a removable discontinuity (e.g., $f(x) = \frac{x^2-1}{x-1}$ with $f(1)$ undefined or defined differently) is still integrable on a closed interval. Riemann integrability is insensitive to changes at finitely many points. As long as the function is bounded and has only finitely many discontinuities, it is integrable. The value at a single point does not affect the limit of Riemann sums, since the contribution of that point to the sum is $f(x_k^*)\Delta x$, which tends to 0 as the mesh size goes to 0.

📖 Explanation: The Thomae function (also known as the popcorn function) is continuous at every irrational number and discontinuous at every rational number. The set of rationals in $[0,1]$ has measure zero. Since the function is bounded (between 0 and 1) and its discontinuities form a set of measure zero, the Lebesgue criterion for Riemann integrability guarantees that it is integrable. Its integral over $[0,1]$ is 0, as the function is nonzero only on a countable set. This function is a standard counterexample to the misconception that integrability requires continuity almost everywhere in the intuitive sense.

📖 Explanation: The Thomae function (option C) is integrable on $[0,1]$ because it is bounded and its discontinuities (the rationals) form a set of measure zero. Option A is unbounded near 0, option B (the Dirichlet function) is discontinuous everywhere, and option D is also unbounded near 0. Both A and D fail the necessary boundedness condition for Riemann integrability. Only the Thomae function satisfies all the criteria for Riemann integrability: boundedness and having discontinuities on a set of measure zero.

📖 Explanation: The theorem about finite discontinuities requires that the function have only finitely many points of discontinuity. Here, the function is discontinuous at every real number in $[-1,1]$: at rational points, the value is 1, but nearby irrational points have value -1, so no limit exists. There are infinitely many discontinuities, and they are not isolated—they form a dense set. The theorem does not apply, and the function fails to be Riemann integrable because the upper and lower Darboux sums never converge to the same value, regardless of how fine the partition is.

📖 Explanation: A bounded function with a single point of discontinuity is always Riemann integrable. The value of the function at that single point does not affect the integral because the contribution of that point to any Riemann sum is $f(c)\Delta x$ for at most one subinterval, and as the mesh size goes to 0, this contribution vanishes. Even if the function is not defined at $c$, it can be assigned any value, and the integral remains unchanged. This is a standard result: finite discontinuities do not affect Riemann integrability provided the function is bounded.

📖 Explanation: A countable set has measure zero, so by the Lebesgue criterion for Riemann integrability, if $f$ is bounded and its discontinuities form a set of measure zero, then $f$ is Riemann integrable. However, boundedness is essential; the criterion applies only to bounded functions. For example, $f(x) = 1/(x - 1/2)$ on $[0,1]$ has a discontinuity at a single point $1/2$, which is countable, but it is unbounded and hence not integrable. So the correct condition is: bounded and discontinuities form a set of measure zero.

📖 Explanation: The behavior of such a function depends entirely on its values. The Thomae function is discontinuous at every rational and continuous at every irrational, and it is integrable because it is bounded and the discontinuities form a measure-zero set. However, if the function takes different bounded values on rationals and irrationals (e.g., 1 on rationals and 0 on irrationals), it is discontinuous everywhere and not integrable. Thus, the mere pattern of discontinuities (rational vs. irrational) is insufficient; the actual values and boundedness determine integrability. The correct answer emphasizes that more information is needed.

📖 Explanation: The error is that the Fundamental Theorem of Calculus requires the integrand to be continuous on the entire closed interval $[-1,1]$. The function $f(x) = 1/x^2$ has an infinite discontinuity (vertical asymptote) at $x = 0$, where it is not defined and not continuous. Additionally, $f$ is unbounded on this interval, which violates the boundedness condition for Riemann integrability. The antiderivative $-1/x$ is also undefined at $x = 0$, so applying the FTC directly is invalid. This would be an improper integral that diverges to infinity, not a finite value.

📖 Explanation: The Thomae function (option B) is the classic example of a function that is integrable despite having infinitely many discontinuities. It is discontinuous at every rational number (infinitely many) but continuous at every irrational number. Since it is bounded (values between 0 and 1) and its discontinuities form a countable set of measure zero, it is Riemann integrable. Options A and C are the Dirichlet function (or variations) which are discontinuous everywhere and not integrable. Option D is unbounded near 0 and thus not integrable.

📖 Explanation: An accumulation point of discontinuities does not automatically prevent integrability. The key condition is the measure of the set of discontinuities. Even if discontinuities accumulate at a point (e.g., $f(x) = \begin{cases} 1, & x = 1/n \\ 0, & \text{otherwise} \end{cases}$ on [0,1]), the discontinuities form a countable set, which has measure zero. Such a function is bounded and integrable. The measure-zero criterion covers both finite and countable sets; it is only when the discontinuities have positive measure that integrability fails. Thus, the presence of accumulation points is not a barrier to integrability.

📖 Explanation: Boundedness is a necessary condition for Riemann integrability; an unbounded function cannot be integrable in the Riemann sense. Continuity is sufficient but not necessary—functions with discontinuities (even infinitely many, as long as they have measure zero) can be integrable. Having finitely many discontinuities is sufficient but not necessary; functions like the Thomae function are integrable with infinitely many discontinuities. Monotonicity is sufficient but not necessary. The only condition that must hold for every Riemann integrable function is boundedness, as the definition of the Riemann integral requires the function to be bounded.

📖 Explanation: A function with a vertical asymptote is unbounded on any interval containing the asymptote. On $[0,1]$, the asymptote at 2 is not included, so the function may be bounded and integrable on that interval. On $[1,2]$, the interval includes the point 2, so the function is unbounded there and not integrable. On $[3,4]$, the function has only a removable discontinuity at 3; such a discontinuity does not affect integrability if the function is bounded on that interval. Therefore, the function is integrable on $[0,1]$ and $[3,4]$ but not on $[1,2]$. The correct answer is the union of intervals where the function is bounded, but the option 'All of the above' is incorrect because $[1,2]$ fails.

📖 Explanation: The set of discontinuities $\{1, 1/2, 1/3, \ldots \}$ is countable and thus has measure zero. However, Riemann integrability also requires the function to be bounded on the interval. If $f$ is bounded, then by the Lebesgue criterion, it is integrable. But without boundedness, the function may fail to be integrable. For example, if $f(1/n) = n$ (unbounded near 0), the function is not integrable despite having countably many discontinuities. Thus, the necessary and sufficient condition is boundedness plus discontinuities of measure zero, so the correct answer emphasizes the boundedness requirement.

📖 Explanation: The statement 'Every bounded function with finitely many discontinuities is Riemann integrable' is actually a true theorem. A counterexample cannot exist because the theorem has been proven: a bounded function with finitely many discontinuities on a closed interval is always Riemann integrable. The Dirichlet function has infinitely many discontinuities (indeed, every point). The function with a single removable discontinuity at 0 is integrable. The function $1/x$ is unbounded, so it violates the boundedness condition and is not a counterexample to a theorem that assumes boundedness.

📖 Explanation: The Lebesgue criterion for Riemann integrability states that a bounded function is Riemann integrable if and only if its set of discontinuities has measure zero. This is a necessary and sufficient condition. The set does not need to be finite or even countable; it can be uncountable as long as its measure is zero (e.g., the Cantor set). It also does not need to be nowhere dense, though many measure-zero sets are. The correct answer is that the discontinuity set must have measure zero, which is the precise characterization of Riemann integrability for bounded functions.

📖 Explanation: The Thomae function is Riemann integrable and its integral over $[0,1]$ is 0. This is because the function is zero at all irrational points, and the rational points where it takes positive values form a countable set. Since countable sets have measure zero, the integral is determined solely by the values on the irrationals, which are 0. Thus, the integral is 0. This function is a classic example of a function that is integrable with infinitely many discontinuities (at the rationals) and whose integral is 0 despite being nonzero on a dense set.

📖 Explanation: The function $f(x) = 1/(x-1)$ has a vertical asymptote at $x = 1$, so it is unbounded on $[0,2]$. The value $f(1) = 5$ does not remove the unboundedness because as $x$ approaches 1 from either side, $|f(x)|$ tends to infinity. Unboundedness is a fatal flaw for Riemann integrability; no bounded Riemann sum can approximate the area under such a function. The discontinuity at 1 is not removable because the limit does not exist (it is infinite). Therefore, the function is not integrable on any interval containing 1.

📖 Explanation: The student's claim is false. The function in option A has a single discontinuity at $x = 0$ (or a removable discontinuity if redefined) and is bounded on $[-1,1]$. Such a function is Riemann integrable; its integral is 0, as the point at 0 contributes nothing to the area. Option B is unbounded and not integrable. Options C and D are the Dirichlet function variants, which are discontinuous everywhere and not integrable. Thus, option A is the correct counterexample to the student's misconception that a single discontinuity prevents integrability.

📖 Explanation: Continuity is a sufficient condition for Riemann integrability, but it is not necessary. Many integrable functions have discontinuities (e.g., step functions with finitely many jumps, the Thomae function). The set of discontinuities having measure zero is both necessary and sufficient (the Lebesgue criterion). Boundedness is necessary but not sufficient (e.g., the Dirichlet function is bounded but not integrable). The discontinuities forming a countable set is sufficient but not necessary? Actually, countable sets have measure zero, so it is sufficient, but not necessary since uncountable measure-zero sets (e.g., Cantor set) are also allowed. Continuity is the strongest unnecessary condition among the options.

📖 Explanation: The pattern of discontinuities alone does not determine integrability; the actual values of the function matter. If $f$ is the Thomae function (values at rationals are $1/q$ and 0 at irrationals), it is integrable. But if $f$ takes different bounded values on rationals and irrationals (e.g., 1 on rationals, 0 on irrationals), it is discontinuous at every point (since every rational is surrounded by irrationals with different values) and not integrable. Both functions have the same discontinuity set (the rationals), yet one is integrable and the other is not. Thus, the values of the function are crucial, not just the set of discontinuities.