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If I define a function such that $m(x)=\int_c^df(x,v)dv$ how can I approach to show continuity in $[a,b]$ if $f$ is continuous in $R=[a,b]\times[c,d]$?

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Hint:

$$|m(x) - m(y)| = \left|\int_c^d (f(x,v) - f(y,v)) \, dv \right| \leqslant \int_c^d|f(x,v) - f(y,v)| \, dv$$

Since $f$ is continuous we can make $|f(x,v) - f(y,v)|$ small when ...

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    Here you are saying that because the uniform continuity on f, because it is continuous and its domain is compact, then I could apply the inequality $|f(x,v)-f(y,v)|<\dfrac{\epsilon}{(d-c)}$ for every $y\in [c,d]$ and get that $\int_c^d |f(x,v)-f(y,v)|<\epsilon$? – 2017-03-04
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    @Mounice: Correct! In fact for every $x,y \in [c,d]$ with $|x-y|$ less than some $\delta(\epsilon)$ your conclusion follows. – 2017-03-04
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    Wait but im saying that for all $\epsilon$ exist $\delta(\epsilon)$ such for $|x-w|<\delta (\epsilon)$ with $x,w\in [a,b]$ then $|f(x,u)-f(w,u)|<\epsilon$ where am I wrong? – 2017-03-06