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In an attempt to prove something about uniform convergence, I need to understand how

$$(1) \hspace{0.5cm} \Big| \int_a^bf(x)-g(x) \,dx \Big| \leq \int_a^b |f(x)-g(x)| \, dx$$

I know it has something to do with the fact that

$$(2) \hspace{0.5cm} -|f(x)-g(x)| \leq f(x)-g(x) \leq |f(x) - g(x)|$$

Another user who posted Integral Inequality Absolute Value: $\left| \int_{a}^{b} f(x) g(x) \ dx \right| \leq \int_{a}^{b} |f(x)|\cdot |g(x)| \ dx$ had a similar question, but theirs dealt with the product. I have already read their post and cannot understand how to carry it over to the difference of two functions.

If we look at the left hand side of my $(1)$, I understand it like this: on the LHS, a potentially negative area becomes positive. On the RHS, a potentially "negative" function (speaking loosely) becomes positive and then the area is taken. Aren't these the same? If not, what's a counterexample? Or a quick small proof.

P.S. I have also read Integral absolute value proof, so you know in advance. But the proof given there doesn't provide an adequate explanation

I think I figured it out

Attempt

$$-|f(x)| \leq f(x) \leq |f(x)|$$

$$-\int_a^b |f(x)| \leq \int_a^b f(x) \leq \int_a^b |f(x)|$$

$$\int_a^b f(x) \leq \int_a^b |f(x)|$$

$$\Big| \int_a^b f(x) \Big| \leq \Big| \int_a^b |f(x)|\Big|$$

But since $|f(x)|$ is already a "positive function" and its area is therefore nonnegative, further instances of taking the absolute value are redundant and thus we can take them away, yielding:

$$\Big| \int_a^b f(x) \Big| \leq \int_a^b |f(x)|$$

SunRoad2
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    https://math.stackexchange.com/a/3031034/42969 (to which you linked) does provide an adequate explanation, in my opinion. – Martin R Sep 11 '21 at 16:00
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    Intuitively, if the function is always positive then you have equality between the two, if instead $f(x)$ has at least an interval in which it is negative it will become a negative contribution on the overall value, if you take the absolute value of $f(x)$ then you make that contribution positive. This is essentially what you have done with your proof. – Tortar Sep 11 '21 at 16:12
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    Because you used a formal approach, in the last passage I would say $|f(x)| \geq 0 \implies \int |f(x)| \geq 0 \implies \int |f(x)| = |\int |f(x)| | $ – Tortar Sep 11 '21 at 16:15

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