Monotonicity of the riemann integral

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The discussion confirms that for integrable functions \( f \) and \( g \) defined on the interval \([a,b]\) with \( f(x) < g(x) \) for all \( x \in [a,b] \), the strict inequality \(\int_a^b f(x)\,dx < \int_a^b g(x)\,dx\) holds true. This is established based on the properties of Riemann integrals, particularly that the integral of a positive function is positive. The discussion also notes that the strict inequality does not hold if \( a = b \), as the integral would equal zero in that case.

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Markjdb
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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)&lt;g(x) for all x\in\left[a,b\right], then do we get the following inequality?
\[\int_a^b f(x)\,dx &lt; \int_a^b g(x)\,dx\]

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.
 
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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.
 

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