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Thread: Integral proof

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    Integral proof

    hello could someone help me with this problem i have an ans but i dont think its worked fully.thanks
    Prove that if f and g are integrable functions on [a,b] then f+g is integrable and

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  2. #2
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    Quote Originally Posted by matty888 View Post
    hello could someone help me with this problem i have an ans but i dont think its worked fully.thanks
    Prove that if f and g are integrable functions on [a,b] then f+g is integrable and

    Let $\displaystyle P(x_0, x_1, ..., x_n)$ be a partition of $\displaystyle [a,b]$ with $\displaystyle I_k=[x_{k-1},x_k]$.

    recall that $\displaystyle sup(f+g) \leq sup \, f + sup \, g$ and $\displaystyle inf(f+g) \geq inf \, f + inf \, g$ over the interval.

    now, for any partition $\displaystyle P$,
    $\displaystyle L(P,f) + L(P,g) \leq L(P, f+g) = \sum_{k=1} ^n inf(f+g)(x_k-x_{k-1})$ ---- (1)

    $\displaystyle U(P,f) + U(P,g) \geq U(P, f+g) = \sum_{k=1} ^n sup(f+g)(x_k-x_{k-1})$ ---- (2)

    So, let $\displaystyle \varepsilon > 0$. since $\displaystyle f$ and $\displaystyle g$ are integrable on $\displaystyle [a,b]$,

    $\displaystyle \exists \, P_{f,\varepsilon}, \, P_{g,\varepsilon}$ such that

    $\displaystyle U(P_{f,\varepsilon},f) - L(P_{f,\varepsilon},f) < \frac{\varepsilon}{2}$ --- (3)

    $\displaystyle U(P_{g,\varepsilon},f) - L(P_{g,\varepsilon},f) < \frac{\varepsilon}{2}$ --- (4)

    so, take $\displaystyle P_{\varepsilon} = P_{f,\varepsilon} \cup P_{g,\varepsilon}$ which is a refinement of $\displaystyle P_{f,\varepsilon}, \, P_{g,\varepsilon}$..


    $\displaystyle U(P_{\varepsilon}, f+g) \leq U(P_{\varepsilon}, f) + U(P_{\varepsilon}, g)$ by (2)

    $\displaystyle \leq U(P_{f,\varepsilon},f) + U(P_{g,\varepsilon},g)$ (from theorem or lemma)

    $\displaystyle \leq L(P_{f,\varepsilon},f) + L(P_{g,\varepsilon},g) + \varepsilon$ by (3) and (4)

    $\displaystyle \leq L(P_{\varepsilon},f) + L(P_{\varepsilon},g) + \varepsilon$ (from theorem or lemma)

    $\displaystyle \leq L(P_{\varepsilon},f+g) + \varepsilon$ by (1)

    and so, $\displaystyle U(P_{\varepsilon},f+g) - L(P_{\varepsilon},f+g) \leq \varepsilon$ which means $\displaystyle f+g$ is integrable on $\displaystyle [a,b]$..

    for the second part, you can do it by showing one is greater or equal than the other and vice versa..
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