Showing the 'integral metric' satisfies the triangle inequality axiom of a metric If this question has already been asked somewhere please let me know and I will close my question, I tried searching for a while but was not sure what to search since I am not sure if 'integral metric' is actually the common name for this metric, or if it even has one. I did find this, but it did not answer my question.
Define $d$ as 
$$d: C[0,1]^2 \to \mathbb{R}, \hspace{3mm}d: (f,g) \mapsto \int_0^1 |f(x)-g(x)| \,dx.$$
I am interested in showing that this function satisfies the triangle inequality axiom of a metric ($d(x,y) \leq d(x,z) + d(z,y)$ for all $x,y,z \in C[0,1]$).
How I proceeded in showing this was:
Let $\, f,g,h \in C[0,1]$. Consider 
$$\int_0^1 |f(x)-g(x)| \,dx + \int_0^1 |g(x)-h(x)| \, dx$$
$$= \int_0^1 |f(x)-g(x)| + |g(x)-h(x)| \,dx.$$
Now by the triangle inequality for real numbers (which we accept without proof) we know that the following statement must hold for any $x \in [0,1]$
$$ |f(x)-g(x)+ g(x)-h(x)| = |f(x)-h(x)| \leq |f(x)-g(x)| + |g(x)-h(x)|.$$
$$\implies \int_0^1 |f(x)-h(x)| \leq \int_0^1 |f(x)-g(x)| + |g(x)-h(x)|,$$
which completes the proof.

My question is if you feel my proof is acceptable. When I was thinking it in my head, I wasn't fully convinced but now that I actually typed it up here I actually feel it might be fine. Particularly, the parts that were bothering me were applying the triangle inequality to the integrand (mainly because $x$ is variable) and the final implication, which seems intuitive, but not fully justified? Thank you for any comments, critiques or discussion.

 A: Your proof is correct. To elaborate on the parts that were causing you discomfort, I'll say the following. See that it holds to apply the triangle inequality to the integrand because $f,g,h$ all map into $\mathbb{R}$, so for all $x \in [0,1]$, let $f(x)-h(x)$ = $a_x$ and $h(x)-g(x)$ = $b_x$, $a_x, b_x \in \mathbb{R}$. Then substituting these quantities for what you have written in the beginning of your proof we have 
$$\int_0^1|a_x| \, dx + \int_0^1 |b_x| \, dx$$
$$= \int_0^1 |a_x| + |b_x| \, dx.$$
Now when we apply the triangle inequality to the integrand it looks more familiar to how we usually see the triangle inequality, and now you might feel more comfortable saying 
$$|a_x+b_x| \leq |a_x| + |b_x|.$$
For the second part you mentioned that seemed unjustified, consider the following lemma. It is a known theorem of calculus that if 
$$h(x) \in C[a,b], h(x) \geq 0 \text{ then } \int_a^b h(x) \, dx \geq 0.$$
Since sum of two elements of $C[a,b]$ is in $C[a,b]$ the theorem implies
$$h(x)-g(x) \geq 0 \implies \int_a^b h(x) - g(x) \, dx \geq 0$$
which applying linearity to the right hand side leads us to the statement that you had reservations about, namely
$$h(x) \geq g(x) \implies \int_a^b h(x) \, dx \geq \int_a^b g(x) \, dx.$$
Applying this lemma to where we left off on the first issue we get the desired result. I hope this makes you more convinced of your proof!
