Integral $=0$ implies function$=0$

Question

Let $f:[a,b]\to\mathbb{R}$ be a mesurable function. How can we show that if $$\int_a ^xf(s)ds=0,$$ for all $x\in[a,b]$, then $f=0$.

Answer 1

It's simply not true. All you can deduce is that $f = 0$ almost everywhere.

Answer 2

Interesting Question. I am not too sure of my answer but feel free to correct me.

If $$\int_a ^xf(s)ds=0,$$ for all for all $x\in[a,b]$. If x=a, then $f(s)$ could be any function. But for all $x\in[a,b]$ then $$\int_a ^xf(s)ds=F(x)-F(a)$$, where $$\frac{d}{dx}F(s)=f(s)$$ and $$F(x)=F(a)$$. The difference in antiderivatives must be zero to satisfy the equation and for that the above equation needs to hold. This would be only true for all $$x$$ if $F'(x)=0$ for all $x$. Since $f$ is locally integrable, $F'(x)=f(x)$ almost everywhere. Hence $f(x)=0$ almost everywhere.

But the official proof of this is given by a theorem.

Theorem. If $f$ is integrable on $[a, b]$ and $\int_a^x {f(t){\rm d}t} = 0$ $\forall x \in [a,b]$, then $f = 0$ a.e. on $[a, b]$.

Proof. An open subset $O$ of $[a,b]$ is a countable union of disjoint open intervals $(c_n, d_n)$; hence, $\int_O {f(t){\rm d}t} = \sum\nolimits_{n = 1}^\infty {\int_{c_n }^{d_n } {f(t){\rm d}t} } = 0$.
If $K$ is a closed subset of $[a,b]$, it thus follows that $\int_K {f(t){\rm }{\rm d}t} = 0$ (note that $[a,b]$ is the disjoint union of $K$ and $[a,b]\backslash K$, and that $[a,b]\backslash K$ is open). Next let $E_ + = \{ x \in [a,b]:f(x) > 0\}$ and $E_ - = \{ x \in [a,b]:f(x) < 0\}$. If $\lambda(E_+) > 0$, then there exists some closed set $K \subset E_+$ such that $\lambda(K) > 0$. But $\int_K {f(t){\rm d}t} = 0$, hence $f=0$ a.e. on $K$. This contradiction shows that $\lambda(E_+) = 0$. Similarly, $\lambda(E_-) = 0$. Therefore $f(t)=0$ for all most every $t\in [a,b]$