# Proving (without Fundamental Theorem of Calculus) that the integral of a continuous function $f \geq 0$ is non-decreasing

just as the title says, I'm working on proving that if $$f : \mathbb{R} \to \mathbb{R}$$ is continuous and $$f(x) \geq 0$$ then

$$g(x) = \int_{0}^{x} f(t)dt$$ is non-decreasing.

I know intuitively that if you have a function that's continuous and positive, then either:

$$f(x) = 0$$ in which case $$g(x) = 0$$ or $$\exists c$$ for which $$f(c) > 0$$, then $$g(x) > 0$$.

How should I go about proving this rigorously though? Any tips are appreciated.

• If $f \ge 0$ and $a \le b$, then $\int_a^b f dt \ge 0$ because all of the Riemann sums are non-negative. Therefore $F(x)=\int_a^x fdt$ is non-decreasing because, if $x < y$, then $F(y)-F(x)=\int_x^y fdt \ge 0$. Mar 13 '20 at 1:59
• Continuity is not needed. Also if $f>0$ then the function $F$ is strictly increasing. This is proved by using the fact a Riemann integrable function is continuous at some point in the interval of integration. Mar 13 '20 at 2:53
• Thank all of you guys! Now I get how to use the definitions to make the proof rigorous. Mar 13 '20 at 4:51

If you define the integral in terms of partitions (as done in Rudin's Principles of Mathematical analysis), then for any given $$x,y\in \mathbb{R}$$ with $$y>x>0$$, let $$P=\{0,x_1, \dots, x_n=y\}$$ be any partition of $$[0,y]$$ containing $$x$$. Then $$P\cap [0,x]=: Q=\{0, x_1, \dots, x_m=x\}$$ is a partition of $$[0,x]$$ and so we have $$U(P,f)=\sum_{i=1}^n M_i \Delta x_i=\sum_{i=1}^m M_i \Delta x_i+\sum_{i=m+1}^n M_i \Delta x_i\geq \sum_{i=1}^m M_i \Delta x_i=U(Q,f).$$ You can do the same thing with $$L(P,f)$$. Note that the inequality uses the fact that $$f\geq 0$$ and so $$M_i=\sup_{[x_{i-1},x_i]}f(x)\geq 0$$. This shows $$g(y)=\int_0^y f(t) \;dt\geq \int_0^x f(t) \; dt=g(x)$$
If $$y \ge x$$, then $$g(y) - g(x) = \int_0^y f(t) \, \mathrm{d}t - \int_0^x f(t) \, \mathrm{d}t = \int_x^y f(t) \, \mathrm{d}t.$$ Here, I'd like to use the fact that, if $$f \ge 0$$ is integrable on $$[x, y]$$ (which it is), then $$\int_x^y f(t) \, \mathrm{d}t \ge 0$$. This would imply that $$g(y) - g(x) \ge 0 \implies g(y) \ge g(x)$$ as required.
If you know this result to be true, then simply refer to it. Otherwise, you can prove it with upper/lower sums. If $$f$$ is integrable over $$[x, y]$$ (which it is), then we know that $$L(f) = U(f)$$, where $$L(f)$$ is the supremum of all lower Riemann sums, and $$U(f)$$ is the infimum of the upper Riemann sums, and both are equal to $$\int_x^y f(t) \, \mathrm{d}t$$. But, since $$f(t) \ge 0$$, the rectangles in the lower Riemann sums all have non-negative height, so the lower sums are all $$\ge 0$$. Taking the supremum $$L(f)$$ must therefore also produce a number $$\ge 0$$, hence $$\int_x^y f(t) \, \mathrm{d}t = L(f) \ge 0,$$ as required.
Let $$x. Then $$g(y)-g(x)=\int_x^yf(t)\,\mathrm{d}t\geq 0$$ since $$f\geq 0$$.