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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.

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    $\begingroup$ 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 $F(y)-F(x)=\int_x^y fdt \ge 0$. $\endgroup$ Mar 13 '20 at 1:59
  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Mar 13 '20 at 2:53
  • $\begingroup$ Thank all of you guys! Now I get how to use the definitions to make the proof rigorous. $\endgroup$
    – Zumus
    Mar 13 '20 at 4:51
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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 $$ g(y)=\int_0^y f(t) \;dt\geq \int_0^x f(t) \; dt=g(x) $$

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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.

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Let $x<y$. Then $g(y)-g(x)=\int_x^yf(t)\,\mathrm{d}t\geq 0$ since $f\geq 0$.

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