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I am making this assertion in a proof, but are doubting wether it is true. The proof is about joint cumulative distributions of continuous random variables, but the proof is more about integration than probability.

If \begin{equation} \forall a\in \mathbb{R}, \int_{-\infty}^{a}f(x)dx = \int_{-\infty}^{a}g(x)dx \end{equation}

Where: \begin{equation} f:\mathbb{R} \to \mathbb{R}, g:\mathbb{R} \to \mathbb{R} \end{equation} (such that their integrals are well-defined)

Can we confidently state that? \begin{equation} f(x) = g(x) \end{equation}

SecchiaChiLegge
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  • @julio_es_sui_glace: that question doesn't have any accepted answers and the answers apart from one comment miss the point just like integrals do $\ddot{\smile}$: information about definite integrals over uncountable spaces tell you nothing about pointwise behaviour without additional assumptions. For the present question, the continuity assumption is relevant. – Rob Arthan Dec 18 '24 at 01:16
  • @RobArthan yes I mis clicked the wrong link however This exact question was on the site not long ago – julio_es_sui_glace Dec 18 '24 at 08:48
  • Here https://math.stackexchange.com/questions/5010352/how-to-prove-that-f-0-almost-everywhere-given-vanishing-lebesgue-integrals-o – julio_es_sui_glace Dec 18 '24 at 08:50
  • Or here https://math.stackexchange.com/questions/1385753/let-m-be-lebesgue-measure-and-a-in-r-suppose-that-f-r-to-r-is-integra – julio_es_sui_glace Dec 18 '24 at 08:52
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    @julio_es_sui_glace: thanks for sorting it out. – Rob Arthan Dec 18 '24 at 22:03

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If $f$ and $g$ are continuous, we can use Fundamental Theorem of Calculus. We can define $F(a) = \int_{-\infty}^a f(x) \, dx$ and $G(a) = \int_{-\infty}^a g(x)\, dx$. We have that $F(a) = G(a)$ for every $a$, and so $f(a) = F'(a) = G'(a) = g(a)$ for every $a$. But if we remove the continuous assumption, then the result isn't true.

Consider $f(x) = \begin{cases} 1, x \neq 0 \\ 0, x = 0 \end{cases}$ $\hspace{5mm}$ and $\hspace{5mm}$ $g(x) = 1$. Then we still have that $\int_{-\infty}^a f(x)\, dx = \int_{-\infty}^a g(x)\, dx$ for every $a$, but $f \neq g$.

Edit: After posting this, I see that someone else linked to a similar question. I can delete this answer if it's not considered helpful.

user1438127
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