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a) False. Counter example $\frac{1}{x}$

b) False. Has to be positive by comparison theorem definition (Cant think of a counter example)

c) True by definite integral properties.

d) True by comparison theorem definition

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    I'm sure this is clear in your mind, but if you consider it from the viewpoint of a Reader on Math.SE, it may occur to you that this is not a self-contained presentation of a Question. Please review [ask]. – 2017-02-18
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    b) let $f(x)=-\frac1x$ and $g(x)=\frac1{x^2}$. This is a counter-example, unless they wanted $0\le f(x)\le g(x)$. – 2017-02-18
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    @SimplyBeautifulArt: Does $\int_{x=0}^\infty g(x) \, dx$ really converge, given the lower limit of integration? – 2017-02-18
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    @BrianTung Close enough, you get what I mean. (for the particular counter-example, just shift it over a bit) – 2017-02-18
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    @SimplyBeautifulArt: Yes, you're right that I get what you mean. That comment was for the benefit of the OP, in case they blindly accept the counterexample. – 2017-02-18
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    How is c not true? Isn't it just a constant multiplying into a value. Like for example if it converges to 5 then c is 1. Unless c could be infinity i don't think it would effect it? And for d isnt that just the opposite of the comparison test, I don't really know how to explain it – 2017-02-18
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    @астонвіллаолофмэллбэрг The OP and I are questioning your logic on part $c$. As far as I can see, it is true. – 2017-02-18
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    I put b as false? Is it not false because by comparision theorem they have to be positive? – 2017-02-18
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    @user349557 b is false unless $0\le f(x)$. Then, it is true. – 2017-02-18
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    @SimplyBeautifulArt I'm extremely sorry about my typing mistake. Yes, your answers for $a,b,c$ are fine, and $d$ is a problem, because taking $f \equiv 0$ and $g \equiv 1$ gives a counterexample. – 2017-02-18
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    Oh so comparison theorem doesnt work opposite ways – 2017-02-18
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    Because $0 \le f \le g$, we have $0 \le \int f \le \int g$ The only way $\int f$ can diverge is if $\int g$ gets out of the way by diverging too. But conversely, if $\int g$ diverges, there is nothing that forces $f$ to go along for the ride. $0 \le e^{-x} \le 1$ for $x \ge 0$, $\int_0^\infty 1dx =\infty$, but $\int_0^\infty e^{-x}dx =1$. – 2017-02-18

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