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The following are, I think, the "worst possible" counterexamples in measure theory. I don't have a nice list of properties, and I have a feeling that I'm forgetting a lot. Feel free to add one in.

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

The following are, I think, the "worst possible" counterexamples in measure theory. They would benefit from a nice list of properties -- I have a feeling that I'm forgetting a lot. Feel free to improve!

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

The following are, I think, the "worst possible" counterexamples in measure theory. I don't have a nice list of properties, and I have a feeling that I'm forgetting a lot. Feel free to add one in.

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$, and the corresponding characteristic function $\chi_A$. ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

The following are, I think, the "worst possible" counterexamples in measure theory. I don't have a nice list of properties, and I have a feeling that I'm forgetting a lot. Feel free to add one in.

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$.   ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

The following are, I think, the "worst possible" counterexamples in measure theory. I don't have a nice list of properties, and I have a feeling that I'm forgetting a lot. Feel free to add one in.

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$, and the corresponding characteristic function $\chi_A$. ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. We are done.

The following are, I think, the "worst possible" counterexamples in measure theory. I don't have a nice list of properties, and I have a feeling that I'm forgetting a lot. Feel free to add one in.

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$, and the corresponding characteristic function $\chi_A$. ($\lambda$ is the Lebesgue measure, ^c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let I_n be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I₁. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval I_n minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I₂, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the I_n-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.