Counterexample for $\int_a^b f < \int_a^b g \Rightarrow f(x)≤g(x), \forall x \in [a,b]$

Question

On Wikipedia, I found out that if $f,g$ are any two Riemann integrable functions then $$f(x)< g(x)\, , \ \forall x \in[a,b] \implies \int_a^b f < \int_a^b g$$

I was wondering if the other way around was true? So (luckily) I made a function that had less area under the curve when above the $X$ axis and the same (signed) area when when below the $X$ axis.

Take $$f(x)= \begin{cases} \frac{1}{2}\sin x & , x \in [0,π] \\ \\ \sin x & , x \in (π,2π]\end{cases}$$ and $g(x) = \sin x , \ x\in [0,2π]$ so $$\int_0^{2π} f < \int_0^{2π} g \nRightarrow f(x)<g(x), \forall x \in [0,2π]$$

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Now, I want a counterexample to $$\int_a^b f < \int_a^b g \Rightarrow f(x)≤g(x), \forall x \in [a,b]$$

Answer 1

If you don't require anything about your functions other than Riemann integrability, that's trivial. Let $f(x)=1$ on $(0,1]$ AND $f(0)=5$. Then the integral is $1$. Let $g(x)=100000x$. That integrates to an area of $50000$ but $f(0)>g(0)$

Answer 2

Another counterexample (with continuous functions).

$f(x)=1\quad$ for any $\;x\in[0,3]\;,$

$g(x)=x\quad$ for any $\;x\in[0,3]\;.$

It results that

$\displaystyle\int_0^3 f(x)\,dx=3<\frac92=\int_0^3 g(x)\,dx$

but $\;f(x)>g(x)\;\;$ for any $\;x\in[0,1[\;.$