# If $f$ is an integrable function, g is a simple function, and $|f(x)|\geq |g(x)|$, then $g$ is integrable, proof verification

Want to prove that if $$f$$ is an integrable function, g is a simple function, and $$|f(x)|\geq |g(x)|$$, then $$g$$ is integrable.

My attempt: For simple function $$g=\sum_{i=1}^n \alpha_i \chi_{E_i}$$ to be integrable we must, per definition, have that $$\mu(E_i)<\infty$$ for all $$i$$ for which $$\alpha_i \neq 0$$, where $$\chi$$ is the characteristic function and $$E_i$$ is a measurable set.

It can be shown that $$f$$ is integrable if and only if $$|f|$$ is integrable, which implies that

$$\int |f(x)|d\mu<\infty$$.

Now by assumption

$$\int |\sum_{i=1}^n \alpha_i \chi_{E_i}|d\mu = \int |g(x)|d\mu \leq \int |f(x)|d\mu<\infty$$

and since we have that

$$\int |\sum_{i=1}^n \alpha_i \chi_{E_i}|d\mu = \sum_{i=1}^n |\alpha_i| \int \chi_{E_i}d\mu = \sum_{i=1}^n |\alpha_i| \mu(E_i)$$,

it follows that $$\mu(E_i)<\infty$$. Is this a valid proof? I'm a bit concerned whether $$\int |g(x)|d\mu \leq \int |f(x)|d\mu$$ makes sense considering the integral symbol $$\int$$ is used when $$g(x)$$ is an integrable fucntion, which is what we want to prove.

Your answer is correct. Also, the symbol $$\say{\int}$$ makes sense for any positive measurable function.
Another way of seeing that is by remembering the definition of the integral of a positive, measurable function (in our case $$|f|$$), as the supremum of the integral over all simple functions below of that function. Now also observe that we can write any simple function $$s(x)= \sum_{i=1}^{n} a_i \chi_{A_i}$$ as $$\sum_{j=1}^{m} b_j \chi_{B_j}$$, where now $$B_i \cap B_j =\emptyset$$, for $$i\neq j$$. This representation gives us that $$|s(x)|=\sum_{j=1}^{m} |b_j| \chi_{B_j}$$, i.e. it is also a simple (and now positive) function. In particular for $$g$$, ee have that $$|g|$$ is a simple function, below $$f$$.
Therefore by the definition of $$\int|f|$$ as the supremum mentioned above, we have the result.