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I add my thoughts to the already existing fine answers.

1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.

Let me add some thoughts to the already existing fine answers.

1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.

I add my thoughts to the already existing fine answers.

1. The answer is "no" as follows by combining Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.

I add my thoughts to the already existing fine answers.

1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.

1. The answer "no" also follows by combining Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.

I add my thoughts to the already existing fine answers.

1. The answer is "no" as follows by combining Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.

2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.