# Monotonicity of the riemann integral

• Markjdb
In summary, the strict version of the property of the Riemann integral holds, meaning that if f(x) is strictly less than g(x) for all x in the interval [a,b], then the integral of f(x) will be strictly less than the integral of g(x). This is due to the fact that a Riemann-integrable function is continuous almost everywhere, and if a function is strictly positive, its integral will also be positive.

#### Markjdb

Hi everyone,

For integrable $$f,g:\left[a,b\right]\rightarrow\mathbb{R}$$ with $$f(x)\leq g(x)$$ for all $$x\in\left[a,b\right]$$, it's a basic property of the riemann integral that
$$$\int_a^b f(x)\,dx \leq \int_a^b g(x)\,dx$$$

My question is whether the strict version of this inequality holds, i.e. if we have the same hypotheses as above, except with $$f(x)<g(x)$$ for all $$x\in\left[a,b\right]$$, then do we get the following inequality?
$$$\int_a^b f(x)\,dx < \int_a^b g(x)\,dx$$$

This question arose while trying to solve a rather different problem; I feel like it's not true in general, but I haven't yet come up with a counterexample.

Yes, you get the strict inequality. The integral of a positive function is obviously positive.

If you take a = b then the strict inequality is not true.

Like jg89 pointed out, it holds as long as the lower and the upper limits of integration are not the same.

Preno said:
Yes, you get the strict inequality. The integral of a positive function is obviously positive.

Is this obvious?

A Riemann-integrable function on [a, b] (with a < b) is continuous almost everywhere, so in particular it's continuous at one point; if this ensures that if the function is strictly positive, then its integral is positive.