# Show that there is a step function $g$ over $$Assume f is integrable over [a,b] and \epsilon > 0. Show that there is a step function g over [a,b] for which g(x) \leq f(x) for all x \in [a,b] and \displaystyle \int_{a}^b (f(x)-g(x))dx < \epsilon. I am having trouble coming up with a step function that satisfies the second condition. Given any f(x), it is easy to come up with a step function such that g(x) \leq f(x) for all x \in [a,b]. But how do we deal with the second condition? #### Solutions Collecting From Web of "Show that there is a step function g over$$"

This follows from the definition of Riemann integral:
For given $\epsilon>0$ there exists $\delta>0$ such that for every partition of $[a,b]$ that is finer than $\delta$, the lower and upper Riemann sum for that partition differ by less than $\epsilon$ from $\int_a^bf(x)\,\mathrm dx$, which is between them. Let $g$ be the step function corresponding to the lower Riemann sum. Then $g(x)\le f(x)$ for all $x$ and the lower Riemann sum is just $\int_a^bg(x)\,\mathrm dx$. Hence $\int_a^b(f(x)-g(x))\,\mathrm dx<\epsilon$, as desired.