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I arrived at something that doesn't make sense, but I can't spot what's wrong with the following proof.

$$ f:\mathbb{Q}\rightarrow\mathbb{Q} \ \ \ \ \ \ \ \ \ f_k:\mathbb{Q}\rightarrow\mathbb{Q}$$ $$f(x)=1$$

Let $K:\{k_1,k_2,k_3...\}$ be a countably infinite set, so we can draw a bijective relationship between $K$ and $\mathbb{Q}[0,1]$ $$ f_{k_{i}} (x)= \begin{cases} 1, & \text{if x = }k_i \\ 0, & \text{if x } \neq k_i \end{cases}$$ Therefore $$\sum_{i=1}^\infty f_{k_{i}}(x) =f(x) \ \ \ \ \ \text{for x}\in \text{[0,1]} $$ However, if we integrate both sides $$\sum_{i=1}^\infty \int_0^1f_{k_{i}}(x)dx =\int_0^1f(x)dx$$ As we know that the area under a point is 0, so that $\int_0^1f_{k_{i}}(x)dx=0$ for all $k_i \in K$ And we have $$0=\sum_{i=1}^\infty 0 =\int_0^1f(x)dx = 1$$

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    Why do you say that $\int_0^1 f(x)~dx=1$? – 2017-02-26
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    because $f(x)=1$ – 2017-02-26
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    That doesn't mean that the integral is $1$. :) – 2017-02-26
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    can you explain a little bit more? – 2017-02-26
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    You have to multiply the function with the measure/size. So your integral is $\int_0^1 f(x)~dx=1\times \mu(\Bbb Q \cap [0,1])=1\times 0=0$ – 2017-02-26
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    Thanks, that make sense. But what if $K$ is a uncountably infinite set and $f$ and $f_k$ are from $ \mathbb{R}\rightarrow\mathbb{R}$? – 2017-02-26
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    What about it? What's your question? – 2017-02-26
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    Then $\mu(\mathbb{R}\cap[0,1])=1$ and would $\int_0^1f(x)dx = 1$? – 2017-02-26
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    Would would $\mu(K \cap [0,1])$ be? It's not obvious. It might be measure $0$ it might be measure $1$. Depending on $K$. – 2017-02-26
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    $K$ has the same measure as the reals, so, 1 – 2017-02-26
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    Just because $K$ is uncountable doesn't mean it has the same measure as the reals. The Cantor set is uncountable but has measure $0$ – 2017-02-26
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    Thanks for the information. But here for the argument's sake, let $K$ has the same measure as the reals. – 2017-02-26
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    Then yeah, sure. – 2017-02-26
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    So if we repeat the original steps to our new $f$ and $f_k$ from $\mathbb{R} \rightarrow \mathbb{R}$ and new $K$ that has the same measure as the reals, would it make $0=1$? – 2017-02-26
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    Nope because measure is only countably additive, not uncountably additive as you found out – 2017-02-26

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