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Liccon Work Planner Liccon Work Planner Software Liccon Work Planner Liccon Work Planner Software Liccon Work Planner Dow LICCON, Work Planner, Work Planner Software, Liccon Work Planner Software, Liccon Work Planner Dow, Liccon Work Planner Software Dow, Liccon Work Planner DowQ: Counterexample to integration by substitution: $f=1$ and $g=1$ I am given the following example from an intuitive standpoint, and I’ve been having a hard time conceptualizing it: I am given that $f$ and $g$ are functions from $(a,b)$ to $\mathbb{R}$ and that $f,g$ and $fg$ have integrals equal to $1$ on the interval $(a,b)$. I must show that either $g$ or $f$ has to be $1$ on the interval. I am then given two equations (which I am sure will be very important): $$\int_a^b (fg)=1$$ $$\int_a^b f=1$$ I must then show that either $f$ or $g$ must be $1$ on the interval. I know how to show this by showing that either $f$ or $g$ are not integrable, or that the integrals don’t exist. I have been stuck at this point for a while now. I have been trying to assume that there is some $c$ on the interval such that $f(c)=g(c)=1$, but there has been no useful path forward to get the integrals I’m given equal to $0$. I’ve also been trying to put the problem into the form of the equality $F(x)-F(a)=G(x)-G(a)$, but have not found much progress there either. I would greatly appreciate it if people could please take the time to provide a solution or point me in the right direction. Thanks in advance. A: You cannot prove this by contradiction: If $f,g \geq 0$, then $$\int_a^b f = \int_a^b g = 1 \implies \int_a^b (f-g) = 0,$$ and so