# Square of a function vanishing outside bounded interval implies function is integrable

lebesgue-integralmeasure-theoryreal-analysis

Prove that if $$f$$ is a non-negative function that vanishes outside a bounded interval, then $$\int_{\mathbb{R}}f^{2}d \lambda < \infty \Rightarrow \int_{\mathbb{R}}f d \lambda < \infty$$

Attempt:

Since $$f$$ vanishes outside of a bounded interval, we only need to consider bounded intervals. $$\int_{\mathbb{R}}f^{2}d \lambda = \int f^{2} \chi_{[a,b]}d \lambda$$. By definition of the Lebesgue integral we can rewrite this as $$\int f^{2 +}\chi_{[a,b]}d \lambda – \int f^{2 -}\chi_{[a,b]}d \lambda$$ where $$f^{+}(x) = \max\{f(x),0\}$$ and $$f^{-}(x)=\max\{-f(x),0\}$$.

Evaluating the above, I get $$(b-a)\max\{f^{2}(x),0\} -(b-a)\max\{-f^{2}(x),0\}$$. Now I am not really sure how to proceed (or if I am on the right track?).

EDIT (After Steven's suggestion)： Let $$A=\{x:|f(x)| \leq 1\} B=\{x:|f(x)|>1\}$$. Then $$\int_{\mathbb{R}}f d \lambda = \int_{A}f d \lambda + \int_{B}f d \lambda$$. Perhaps I then split this into positive and negative parts and try to somehow use the condition on $$f^{2}$$?

Given $$f$$ vanishes outside a bounded interval $$A$$.

Now using C-S inequality:

$$\int_A f \cdot 1 \, d\lambda \leqslant \left( \int_A f^2 \, d\lambda \right)\left( \int_A \, d\mu \right) = \lambda(A) \int_A f^2 \, d\lambda$$

$$<\infty$$

Here $$\lambda(A)=\ell(A)< \infty$$ as $$A$$ is bounded interval and $$\int_{A}f^2 d\lambda<\infty$$

Hence $$f\in L^1(A)$$ and $$f=0$$ on $$\Bbb{R}\setminus A$$.

Hence $$f\in L^1(\Bbb{R})$$ i.e $$\int_{\Bbb{R}}fd\lambda<\infty$$