# Lebesgue integrable implies zero almost everywhere

Let $$f$$ be an integrable function on $$\mathbb{R}$$ such that for every continuous $$g$$ on $$\mathbb{R}$$, $$\displaystyle\int_\mathbb{R} fg \ dm=0$$. Show that $$f$$ is zero a.e.

This is a qualifier sample problem and my idea is to proceed by contradiction. My problem here is that $$f$$ is not necessarily nonnegative so I'm having difficulties on establishing inequalities to arrive at a contradiction. Thanks!

• Hint: You can always write a measurable function $f$ as $f =f^++f^-$ where $f^+(x)=\max\{0,f(x)\}$ and $f^-(x)=-\max\{0,-f(x)\}$. So $f\neq 0$ a.e. implies $f^+$ or $f^-$ must be non-zero on some set of positive measure. Apr 26, 2019 at 17:45
• If we write $f$ in that form, I don't see why that works. Apr 26, 2019 at 18:06
• $\int_\mathbb R fg~dm=0$ for every continuous $g$ implies that $\int_I f~dm=0$ over every interval $I$. Let $g$ be a bump with height 1 on $I$ and very steep slopes. If $f$ were to have some oscillatory/cancellation behavior, then it will need to oscillate infinitely fast to do so on every interval. Furthermore there must be a simple function that approximates $f$ on $I$. Such a simple function must be very small on any set of positive measure. Not rigorous, just some thoughts. Apr 26, 2019 at 18:20
• You can cheat; the assumptions imply that the Fourier transform $\hat{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx$ vanishes, and since the Fourier transform is injective, $f=0$. The cheating part is that you should prove that the Fourier transform is injective, of course. Apr 26, 2019 at 18:58
• @GiuseppeNegro LOL. Now we are cheating, we could also say that $f$ is weakly differentiable with weak derivative $0$, so it must be constant a.e.. But then the only integrable constant is zero. Here we did all kind of cheating (and circle reasoning)... Apr 26, 2019 at 19:03

Hint: If $$b\in L^\infty(\mathbb R),$$ then there exists a sequence $$g_n$$ of bounded continuous functions on $$\mathbb R$$ such that $$\|g_n\|_\infty\le \|b\|_\infty$$ for all $$n,$$ and $$g_n \to b$$ pointwise a.e.
$$\newcommand{\sign}{\operatorname{sign}}$$By taking continuous functions $$g$$ with support in $$(0,1)$$, we can conclude that the statement is true for all functions $$g\in C_c((0,1))$$, i.e. functions with compact support in $$(0,1)$$, right? So let us focus on the function $$f|_{(0,1)}$$ on $$(0,1)$$ and forget about $$\mathbb R$$ for the moment.
Since $$(0,1)$$ equipped with the Lebesgue measure $$m$$ is a finite measure space, Lusin tells us that for every $$\varepsilon>0$$ there exists a compact set $$K_\varepsilon\subset (0,1)$$ such that $$m((0,1)\setminus K_\varepsilon)<\varepsilon$$ and a continuous function $$h_\varepsilon$$ with compact support such that $$h_\varepsilon=\sign(f)$$ on $$K_\varepsilon$$ and $$\sup_{x\in (0,1)}|h_\varepsilon(x) |\leq 1$$. This means for instance that we can take $$g=h_\varepsilon$$ and get $$0=\int_{(0,1)}fg\,dm=\int_{K_\varepsilon} |f|\,dm + \int_{(0,1)\setminus K_\varepsilon}fh_{\varepsilon}\,dm$$ Now we take a sequence of $$\varepsilon$$s, say $$\varepsilon_n=1/n$$, and apply DCT (HOW?) to get that $$\int_{(0,1)}|f|\,dm=0$$ which implies $$f=0$$ a.e. on $$(0,1)$$. But we could also take $$(a,b)$$ instead of $$(0,1)$$, right? Of course, try it yourself and see if you understand it.
• Edited. I have added a crucial thing about $h_\varepsilon$. May 5, 2019 at 14:08