We have the functions $f$ and $g$ such that,
$$f:\mathbb{[0,1]}\rightarrow\mathbb{R}$$
$$g:\mathbb{[0,1]}\rightarrow\mathbb{R}$$
and both $f$ and $g$ are bounded and continuos functions. Show that,
$$\int_0^1 |f(x) + g(x)|\:dx \: \le \int_0^1 |f(x)| \:dx + \int_0^1 |g(x)| \:dx $$
Could anyone help me out on this one ?
On the technicality discussed above:
As $f$ and $g$ are continuous on a closed interval $[0,1]$, the integrals $\int_0^1 f(x) \ dx$, $\int_0^1 g(x) \ dx$ exist by standard results.
Further, $|f|$ is also a continuous function on $[0,1]$ because it is the composite of two continuous functions, $f$ and absolute value $| \cdot |$. Hence $|f|$ is integrable on $[0,1]$. Likewise for $|g|$.
Finally, $|f + g|$ is also continuous, being the composite of the sum of two continuous functions--which is also continuous--and absolute value; thus $\int_0^1 |f(x) + g(x)| \ dx$ also exists.