# real analysis – Prove that \$ left | int_X mathbf {f} d mu right | _p leq int_X left | mathbf {f} right | _p d mu \$ for \$ mathbf {f} = (f_1, …, f_n) \$

I have problems with the following proof.

Define $$mathbf {f} (x) = (f_1 (x), …, f_n (x))$$ Where $$f_i: X to mathbb {R}$$ for every positive integer $$i leq n$$, and each one $$f_i$$ is integrable.

Prove that
$$left | int_X mathbf {f} \ mathrm {d} mu right | _p leq int_X left | mathbf {f} right | _p \ mathrm {d} mu$$

to the $$1 leq p leq infty$$

Now we can explicitly write the left side as

$$left | int_X mathbf {f} \ mathrm {d} mu right | _p = left ( sum_ {j = 1} ^ n left | int_X f_j \ mathrm {d} mu right | ^ p right) ^ {1 / p}$$

Now the exponent annoys me. If we had the case $$p = 1$$, then it's just there for every integrable function $$g: X to mathbb {R}$$, we have

$$left | int_X g \ mathrm {d} mu right | leq int_X | g | \ mathrm {d} mu$$

Consequently,

$$left | int_X mathbf {f} \ mathrm {d} mu right | = sum_ {j = 1} ^ n left | int_X f_j mathrm {d} mu right | leq sum_ {j = 1} ^ n int_X | f_j | \ mathrm {d} mu = int_X left ( sum_ {j = 1} ^ n | f_j | right) \ mathrm {d} mu = int_X | mathbf {f} | \ mathrm {d} mu$$

But I can not handle the exponential $$1

Thanks!