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

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

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

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

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

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

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

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


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

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

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

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

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

    and so, U(P_{\varepsilon},f+g) - L(P_{\varepsilon},f+g) \leq \varepsilon which means f+g is integrable on [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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